Let the random variable \(X\) represent the weight of a large bag of pasta. \(X\) is normally distributed with mean \(\mu = 1.5\) kg and standard deviation \(\sigma = 0.05\) kg.

We need to find \(P(1.42 < X < 1.52)\).

Standardize the variable using the formula:

\(Z = \frac{X - \mu}{\sigma}\)

For \(X = 1.42\):

\(Z = \frac{1.42 - 1.5}{0.05} = -1.6\)

For \(X = 1.52\):

\(Z = \frac{1.52 - 1.5}{0.05} = 0.4\)

Thus, \(P(1.42 < X < 1.52) = P(-1.6 < Z < 0.4)\).

Using standard normal distribution tables or a calculator, find:

\(\Phi(0.4) = 0.6554\)

\(\Phi(-1.6) = 0.0548\)

Therefore, \(P(-1.6 < Z < 0.4) = \Phi(0.4) - \Phi(-1.6) = 0.6554 - 0.0548 = 0.6006\).

Rounding to three decimal places, the probability is \(0.601\).

(a) To find the probability that Raj runs for more than 43.2 minutes, we use the standard normal distribution. First, we calculate the z-score:

\(z = \frac{43.2 - 41.2}{3.6} = 0.5556\)

We need to find \(P(Z > 0.5556)\). Using the standard normal distribution table, \(\Phi(0.5556) = 0.7108\).

Therefore, \(P(Z > 0.5556) = 1 - 0.7108 = 0.2892\).

So, the probability that Raj runs for more than 43.2 minutes is approximately 0.289.

(b) To find the number of days Raj runs for less than 43.2 minutes, we calculate:

\(P(X < 43.2) = 1 - P(X > 43.2) = 1 - 0.2892 = 0.7108\).

Multiply this probability by the number of days in a year:

\(0.7108 \times 365 = 259.4\).

Therefore, the estimated number of days is approximately 259 or 260.

(a) To find the probability that a randomly chosen employee takes more than 28.6 minutes, we use the standard normal distribution. First, we calculate the z-score:

\(z = \frac{28.6 - 32.2}{9.6} = -0.375\)

We need \(P(Z > -0.375)\). Using the standard normal distribution table, \(\Phi(-0.375) = 0.3538\). Therefore, \(P(Z > -0.375) = 1 - 0.3538 = 0.6462\).

Thus, the probability is approximately 0.646.

(c) We need to find the probability that the time differs from the mean by less than 15.0 minutes, i.e., \(|X - 32.2| < 15\).

This is equivalent to \(17.2 < X < 47.2\). We calculate the z-scores:

\(z_1 = \frac{17.2 - 32.2}{9.6} = -1.5625\)

\(z_2 = \frac{47.2 - 32.2}{9.6} = 1.5625\)

We need \(P(-1.5625 < Z < 1.5625)\). Using the standard normal distribution table, \(\Phi(1.5625) = 0.9409\).

Thus, \(P(-1.5625 < Z < 1.5625) = 2 \times 0.9409 - 1 = 0.8818\).

Therefore, the probability is approximately 0.882.

(i) To find the number of days Karli spends more than 142 minutes on social media, we first calculate the probability that she spends more than 142 minutes on a given day. We use the standard normal distribution:

\(P(X > 142) = P\left( Z > \frac{142 - 125}{24} \right) = P(Z > 0.7083)\)

Using the standard normal distribution table, \(P(Z > 0.7083) = 1 - 0.7604 = 0.2396\).

Therefore, the expected number of days is \(0.2396 \times 365 = 87.454\).

Rounding to the nearest whole number, we get 87 or 88 days.

(ii) We need to find the probability that Karli spends more than 142 minutes on social media on fewer than 2 out of 10 days. This is a binomial probability problem with \(n = 10\) and \(p = 0.2396\).

\(P(0, 1) = 0.7604^{10} + \binom{10}{1} \times 0.2396^1 \times 0.7604^9\)

\(= 0.064628 + 0.20364\)

\(= 0.268\)

We need to find the probability that a rod's length is within 0.5 cm of the mean, i.e., between 24.7 cm and 25.7 cm.

Using the standardization formula, we calculate:

\(P\left( \frac{25.2 - (25.2 + 0.5)}{0.4} < z < \frac{25.2 - (25.2 - 0.5)}{0.4} \right)\)

This simplifies to:

\(P\left( \frac{-0.5}{0.4} < z < \frac{0.5}{0.4} \right) = P(-1.25 < z < 1.25)\)

Using the standard normal distribution table, we find:

\(P(-1.25 < z < 1.25) = 2 \Phi(1.25) - 1\)

\(\Phi(1.25) \approx 0.8944\)

Thus, \(2 \times 0.8944 - 1 = 0.7888\)

The expected number of rods is:

\(0.7888 \times 500 = 394.4\)

Rounding gives 394 or 395 rods.

Let the random variable representing the time spent be denoted by \(X\), where \(X \sim N(96, 18^2)\).

We need to find \(P(85 < X < 100)\).

Standardize the variable: \(Z = \frac{X - 96}{18}\).

Thus, \(P(85 < X < 100) = P\left(\frac{85 - 96}{18} < Z < \frac{100 - 96}{18}\right)\).

Calculate the standardized values: \(\frac{85 - 96}{18} = -0.6111\) and \(\frac{100 - 96}{18} = 0.2222\).

Therefore, \(P(-0.6111 < Z < 0.2222) = \Phi(0.2222) + \Phi(0.6111) - 1\).

Using standard normal distribution tables, \(\Phi(0.2222) = 0.5879\) and \(\Phi(0.6111) = 0.7294\).

Thus, \(P(-0.6111 < Z < 0.2222) = 0.5879 + 0.7294 - 1 = 0.3173\).

Rounding to three decimal places, the probability is \(0.317\).

Let the random variable \(X\) represent the time taken to swim 100 metres. \(X\) follows a normal distribution with mean \(\mu = 62\) and standard deviation \(\sigma = 5\).

We need to find \(P(56 < X < 66)\).

Standardize the variable using the formula:

\(Z = \frac{X - \mu}{\sigma}\)

For \(X = 56\):

\(Z = \frac{56 - 62}{5} = -1.2\)

For \(X = 66\):

\(Z = \frac{66 - 62}{5} = 0.8\)

Thus, \(P(56 < X < 66) = P(-1.2 < Z < 0.8)\).

Using the standard normal distribution table, find:

\(\Phi(0.8) = 0.7881\)

\(\Phi(1.2) = 0.8849\)

Therefore, \(P(-1.2 < Z < 0.8) = \Phi(0.8) + \Phi(1.2) - 1 = 0.7881 + 0.8849 - 1 = 0.673\).

(a) To find the probability that Pia takes longer than 11.3 minutes, we use the standard normal distribution. First, we standardize the variable:

\(P(X > 11.3) = P\left( Z > \frac{11.3 - 10.1}{1.3} \right) = P(Z > 0.9231)\)

Using the standard normal distribution table, \(P(Z > 0.9231) = 1 - P(Z < 0.9231) = 1 - 0.822\)

Thus, \(P(X > 11.3) = 0.178\).

(c) To find the number of days Pia takes between 8.9 and 11.3 minutes, we calculate:

\(P(8.9 < X < 11.3) = 1 - 2 \times P(X > 11.3)\)

\(= 2 \times (0.5 - 0.178) = 0.644\)

The expected number of days is \(90 \times 0.644 = 57.96\), which rounds to 57 days.

(a) To find the probability that Davin plays for more than 4.2 hours, we use the standard normal distribution. First, we standardize the variable:

\(P(X > 4.2) = P\left( Z > \frac{4.2 - 3.5}{0.9} \right) = P(Z > 0.7778)\)

Using the standard normal distribution table, \(P(Z > 0.7778) = 1 - 0.7818 = 0.218\).

(c) To find the number of days Davin plays between 2.8 and 4.2 hours, we calculate:

\(P(2.8 < X < 4.2) = 1 - 2 \times P(X > 4.2)\)

\(= 2 \times (1 - 0.7818) - 1 = 2 \times 0.2182 - 1 = 0.5636\)

The number of days is \(365 \times 0.5636 = 205.7\), so approximately 205 days.

The time, X, is normally distributed with mean \(\mu = 15.8\) hours and standard deviation \(\sigma = 4.2\) hours.

We need to find \(P(X < 21)\).

First, calculate the z-score for 21 hours:

\(z = \frac{21 - 15.8}{4.2} = 1.238\)

Using the standard normal distribution table, find \(\Phi(1.238)\), which gives the probability that a standard normal variable is less than 1.238.

\(\Phi(1.238) = 0.892\)

Thus, \(P(X < 21) = 0.892\).

Let the random variable representing the length of the snakes be denoted by \(X\), where \(X \sim N(54, 6.1^2)\).

We need to find \(P(50 < X < 60)\).

First, convert the lengths to standard normal variables \(Z\) using the formula:

\(Z = \frac{X - \mu}{\sigma}\)

For \(X = 50\):

\(Z = \frac{50 - 54}{6.1} = -0.6557\)

For \(X = 60\):

\(Z = \frac{60 - 54}{6.1} = 0.9836\)

Thus, we need to find \(P(-0.6557 < Z < 0.9836)\).

This is equivalent to \(\Phi(0.9836) - \Phi(-0.6557)\).

Using the standard normal distribution table:

\(\Phi(0.9836) = 0.8375\)

\(\Phi(-0.6557) = 1 - \Phi(0.6557) = 1 - 0.7441 = 0.2559\)

Therefore, \(P(-0.6557 < Z < 0.9836) = 0.8375 - 0.2559 = 0.5816\).

Rounding to three decimal places, the probability is \(0.582\).

To find the expected number of birds with lengths between 15.4 cm and 16.8 cm, we first standardize the lengths using the formula for the standard normal distribution:

\(Z = \frac{X - \mu}{\sigma}\)

For 15.4 cm: \(Z = \frac{15.4 - 16.5}{0.6} = -1.833\)

For 16.8 cm: \(Z = \frac{16.8 - 16.5}{0.6} = 0.5\)

We need to find \(P(-1.833 < Z < 0.5)\).

Using the standard normal distribution table, we find:

\(\Phi(0.5) = 0.6915\)

\(\Phi(-1.833) = 1 - \Phi(1.833) = 1 - 0.9666 = 0.0334\)

Thus, \(P(-1.833 < Z < 0.5) = 0.6915 - 0.0334 = 0.6581\).

The expected number of birds is:

\(0.6581 \times 150 = 98.715\)

Rounding to the nearest whole number, we expect 99 birds.

We need to find the probability that a student's height is within half a standard deviation of the mean, i.e., between 144 cm and 152 cm.

Using the standardization formula, we calculate:

\(P(144 < X < 152) = P\left( \frac{144 - 148}{8} < Z < \frac{152 - 148}{8} \right)\)

\(= P\left( -\frac{1}{2} < Z < \frac{1}{2} \right)\)

Using the standard normal distribution table, we find:

\(P\left( -\frac{1}{2} < Z < \frac{1}{2} \right) = 0.6915 - (1 - 0.6915) = 2 \times 0.6915 - 1\)

\(= 0.383\)

Therefore, the expected number of students is:

\(0.383 \times 120 = 45.96\)

Rounding to the nearest whole number, we accept 45 or 46.

(i) To find the probability that a fir tree has a height less than 45 metres, we standardize the variable using the formula:

\(Z = \frac{X - \mu}{\sigma}\)

where \(X = 45\), \(\mu = 40\), and \(\sigma = 8\).

\(Z = \frac{45 - 40}{8} = 0.625\)

We find \(P(Z < 0.625)\) using standard normal distribution tables or a calculator, which gives:

\(P(X < 45) = 0.734\)

(ii) To find the probability that a fir tree has a height within 5 metres of the mean, we calculate:

\(P(35 < X < 45) = P(X < 45) - P(X < 35)\)

Using the result from part (i), \(P(X < 45) = 0.734\).

Standardize for \(X = 35\):

\(Z = \frac{35 - 40}{8} = -0.625\)

\(P(Z < -0.625) = 1 - P(Z < 0.625) = 1 - 0.734 = 0.266\)

Thus, \(P(35 < X < 45) = 0.734 - 0.266 = 0.468\)

(i) To find the probability that a randomly chosen athlete has a PB between 46 and 53 seconds, we standardize the values using the formula:

\(P(46 < X < 53) = P\left( \frac{46 - 49.2}{2.8} < Z < \frac{53 - 49.2}{2.8} \right)\)

This simplifies to:

\(P(-1.143 < Z < 1.357)\)

Using the standard normal distribution table, we find:

\(\Phi(1.357) + \Phi(1.143) - 1 = 0.9126 + 0.8735 - 1\)

Thus, the probability is:

0.786

(iii) To find the probability that exactly 2 out of 3 athletes have PBs of less than 46 seconds, we first find:

\(P(X < 46) = 0.1265\)

Then, using the binomial probability formula:

\(P(2 \text{ PB} < 46) = 3(1 - 0.1265)0.1265^2\)

This calculates to:

0.0419

Let the time taken be represented by the random variable \(X\), where \(X \sim N(85, 6.8^2)\).

We need to find \(P(79 < X < 91)\).

Standardize the variable using the formula \(Z = \frac{X - \mu}{\sigma}\), where \(\mu = 85\) and \(\sigma = 6.8\).

Calculate the z-scores:

For \(X = 79\), \(Z = \frac{79 - 85}{6.8} = -0.8824\).

For \(X = 91\), \(Z = \frac{91 - 85}{6.8} = 0.8824\).

Thus, \(P(79 < X < 91) = P(-0.8824 < Z < 0.8824)\).

Using the standard normal distribution table, find:

\(\Phi(0.8824) = 0.8111\) and \(\Phi(-0.8824) = 1 - 0.8111 = 0.1889\).

Therefore, \(P(-0.8824 < Z < 0.8824) = 0.8111 - 0.1889 = 0.622\).

We need to find the probability that a cartridge contains less than 28.9 ml of ink. Let the random variable representing the volume of ink be normally distributed as \(X \sim N(30, 1.5^2)\).

We standardize the variable to find \(P(X < 28.9)\):

\(P(X < 28.9) = P\left( Z < \frac{28.9 - 30}{1.5} \right) = P(Z < -0.733)\)

Using the standard normal distribution table, \(P(Z < -0.733) = 1 - 0.7682 = 0.2318\).

The expected number of cartridges per month that contain less than 28.9 ml is:

\(0.2318 \times 8 = 1.85\)

Rounding to the nearest whole number, we get 2 cartridges.

To find the number of giraffes weighing less than 700 kg, we first calculate the z-score for 700 kg using the formula:

\(z = \frac{X - \mu}{\sigma}\)

where \(X = 700\), \(\mu = 830\), and \(\sigma = 120\).

\(z = \frac{700 - 830}{120} = -1.083\)

Next, we find the probability that a giraffe weighs less than 700 kg, \(P(X < 700)\), which is equivalent to \(P(z < -1.083)\).

Using standard normal distribution tables or a calculator, \(P(z < -1.083) = 0.1394\).

Therefore, the expected number of giraffes weighing less than 700 kg is:

\(430 \times 0.1394 = 59.9\)

Rounding to the nearest whole number, we get 59 or 60 giraffes.

Let the random variable \(X\) represent the travel time in minutes. \(X\) is normally distributed with mean \(\mu = 140\) and standard deviation \(\sigma = 12\).

We need to find \(P(X < 132)\).

First, standardize the variable using the formula:

\(Z = \frac{X - \mu}{\sigma}\)

Substitute the values:

\(Z = \frac{132 - 140}{12} = -0.6667\)

Now, find \(P(Z < -0.6667)\).

Using standard normal distribution tables or a calculator, \(P(Z < -0.6667) = 0.252\).

(i) To find the probability that an apple is graded as medium, we need to calculate the probability that its weight is between 90 and 140 grams. We standardize these values using the formula:

\(z = \frac{x - \mu}{\sigma}\)

For 90 grams: \(z_1 = \frac{90 - 120}{24} = -\frac{5}{4}\)

For 140 grams: \(z_2 = \frac{140 - 120}{24} = \frac{5}{6}\)

The probability that an apple is graded as medium is:

\(\Phi\left(\frac{5}{6}\right) - \Phi\left(-\frac{5}{4}\right)\)

Using standard normal distribution tables or a calculator, we find:

\(\Phi(0.8333) - (1 - \Phi(1.25)) = 0.7975 - 0.1056 = 0.6919\)

Rounding to 3 significant figures gives 0.692.

(ii) We use the binomial distribution to find the probability that at least two apples are graded as medium. Let \(X\) be the number of medium apples out of 4, where \(X \sim \text{Binomial}(4, 0.692)\).

We need \(P(X \geq 2) = 1 - P(X = 0) - P(X = 1)\).

\(P(X = 0) = (1 - 0.692)^4\)

\(P(X = 1) = 4 \times 0.692^1 \times (1 - 0.692)^3\)

Calculating these:

\(P(X = 0) = 0.00899\)

\(P(X = 1) = 0.0808757\)

Thus, \(P(X \geq 2) = 1 - 0.00899 - 0.0808757 = 0.910\).

We need to find \(P(Y > 0)\). Since \(Y\) is normally distributed with mean \(\mu\) and standard deviation \(\sigma\), we can standardize \(Y\) to a standard normal variable \(Z\) using the formula:

\(Z = \frac{Y - \mu}{\sigma}\)

Thus, \(P(Y > 0) = P\left( Z > \frac{0 - \mu}{\sigma} \right)\).

Given \(4\sigma = 3\mu\), we can express \(\mu\) in terms of \(\sigma\):

\(\mu = \frac{4\sigma}{3}\)

Substitute \(\mu\) into the standardization formula:

\(P\left( Z > \frac{0 - \frac{4\sigma}{3}}{\sigma} \right) = P\left( Z > \frac{-4\sigma/3}{\sigma} \right) = P\left( Z > -\frac{4}{3} \right)\)

Using the standard normal distribution table, \(P(Z > -4/3) = 0.909\).

To find the probability that \(X < 29.4\), we first standardize the variable using the formula for the standard normal distribution:

\(P(X < 29.4) = P\left(Z < \frac{29.4 - 31.4}{\sqrt{3.6}}\right)\)

Calculating the standardized value:

\(\frac{29.4 - 31.4}{\sqrt{3.6}} = -1.0541\)

Thus, \(P(X < 29.4) = P(Z < -1.0541)\).

Using the standard normal distribution table, we find:

\(P(Z < -1.0541) = 1 - 0.8540 = 0.146\)

Therefore, the probability that a randomly chosen value of \(X\) is less than 29.4 is 0.146.

First, we standardize the time of 20 minutes using the formula for the standard normal distribution:

\(Z = \frac{X - \mu}{\sigma}\)

where \(X = 20\), \(\mu = 14.6\), and \(\sigma = 5.2\).

\(Z = \frac{20 - 14.6}{5.2} = 1.03846\)

Next, we find the probability that \(Z > 1.03846\). Using standard normal distribution tables or a calculator, we find:

\(P(Z > 1.03846) = 1 - P(Z < 1.03846) = 1 - 0.8504 = 0.1496\)

This means approximately 14.96% of students take more than 20 minutes.

In a sample of 250 students, the expected number is:

\(250 \times 0.1496 = 37.4\)

Rounding to the nearest whole number, we expect about 37 or 38 students to take more than 20 minutes.

We start by standardizing the variable. We want to find \(P(X < 3\mu)\).

Standardizing, we have:

\(P(X < 3\mu) = P\left( z < \frac{3\mu - \mu}{4\mu/3} \right)\)

\(= P\left( z < \frac{9\sigma/4 - 3\sigma/4}{\sigma} \right)\)

\(= P\left( z < \frac{6}{4} \right)\)

Using the standard normal distribution table, \(P(z < 1.5) = 0.933\).

To find \(P(X > 0)\), we standardize the variable using the formula for the standard normal distribution:

\(P(X > 0) = P\left( z > \frac{0 - \mu}{\sigma} \right)\)

Substitute \(\mu = 1.5\sigma\):

\(P\left( z > \frac{0 - 1.5\sigma}{\sigma} \right) = P(z > -1.5)\)

Using the standard normal distribution table, \(P(z > -1.5) = 0.933\).

First, we standardize the lengths using the formula for the standard normal variable:

\(z = \frac{x - \mu}{\sigma}\)

For rods shorter than 15.75 cm:

\(P(x < 15.75) = P\left(z < \frac{15.75 - 16}{0.2}\right) = P(z < -1.25)\)

Using symmetry and standard normal tables, \(P(z < -1.25) = 1 - P(z < 1.25) = 1 - 0.8944 = 0.1056\).

For rods longer than 16.25 cm:

\(P(x > 16.25) = P\left(z > \frac{16.25 - 16}{0.2}\right) = P(z > 1.25) = 0.1056\) (by symmetry).

The probability of a rod being usable is:

\(P(\text{usable}) = 1 - P(x < 15.75) - P(x > 16.25) = 1 - 0.1056 - 0.1056 = 0.7888\).

The expected number of usable rods in a batch of 1000 is:

\(1000 \times 0.7888 = 788.8\).

Rounding gives approximately 788 or 789 usable rods.

Let the mean and standard deviation of the normal distribution be denoted by \(\mu\). Since the mean is equal to the standard deviation, we have \(\sigma = \mu\).

We need to find \(P(X < 1.5\mu)\).

Standardizing the variable, we use the formula:

\(z = \frac{X - \mu}{\sigma}\)

Substituting \(X = 1.5\mu\), \(\mu = \sigma\), we get:

\(z = \frac{1.5\mu - \mu}{\mu} = \frac{0.5\mu}{\mu} = 0.5\)

Thus, \(P(X < 1.5\mu) = P(z < 0.5)\).

Using standard normal distribution tables or a calculator, \(P(z < 0.5) = 0.692\).

First, we standardize the normal variable to find the probability that a single woman's middle finger is shorter than 8.2 cm:

\(P(X < 8.2) = P\left( Z < \frac{8.2 - 7.9}{0.44} \right) = P(Z < 0.6818)\)

Using the standard normal distribution table, \(P(Z < 0.6818) \approx 0.7524\).

Now, we use the binomial distribution to find the probability that exactly 3 out of 5 women have middle fingers shorter than 8.2 cm:

\(P(3) = \binom{5}{3} (0.7524)^3 (1 - 0.7524)^2\)

\(P(3) = 10 \times (0.7524)^3 \times (0.2476)^2\)

\(P(3) \approx 0.261\)

Let the random variable for the time taken to cook an egg be denoted as \(X\), where \(X \sim N(4.2, 0.6^2)\).

We need to find \(P(3.5 < X < 4.5)\).

Standardize the variable using the formula for the standard normal distribution: \(Z = \frac{X - \mu}{\sigma}\).

For \(X = 4.5\):

\(Z = \frac{4.5 - 4.2}{0.6} = \frac{0.3}{0.6} = 0.5\).

Thus, \(P(X < 4.5) = P(Z < 0.5) = 0.6915\).

For \(X = 3.5\):

\(Z = \frac{3.5 - 4.2}{0.6} = \frac{-0.7}{0.6} = -1.167\).

Thus, \(P(X < 3.5) = P(Z < -1.167) = 0.1216\).

Therefore, \(P(3.5 < X < 4.5) = P(X < 4.5) - P(X < 3.5) = 0.6915 - 0.1216 = 0.570\).

(i) To find the probability that Peter takes longer than 10.2 minutes, we standardize the variable:

\(P(x > 10.2) = P\left( z > \frac{10.2 - 9.5}{1.3} \right)\)

\(= P(z > 0.53846)\)

Using the standard normal distribution table, \(P(z > 0.53846) = 1 - 0.7046 = 0.295\).

(ii) To find the number of days Peter takes less than 8.8 minutes, we use symmetry:

\(P(x < 8.8) = 0.2954\)

Estimate the number of days: \(365 \times 0.2954 = 107.671\)

Rounding gives approximately 107 or 108 days.

Given that the times follow a normal distribution \(N(\mu, \sigma^2)\) with \(\mu = 3\sigma\), we need to find \(P(t > 0.6\mu)\).

First, standardize the variable:

\(P(t > 0.6\mu) = P\left( z > \frac{0.6\mu - \mu}{\mu/3} \right)\)

Substitute \(\mu = 3\sigma\) into the equation:

\(P\left( z > \frac{0.6 \times 3\sigma - 3\sigma}{3\sigma/3} \right) = P(z > -1.2)\)

Using the standard normal distribution table, \(P(z > -1.2) = 0.885\).

We need to find \(P(Y < 2\mu)\). To do this, we standardize the variable using the formula for the standard normal distribution:

\(z = \frac{Y - \mu}{\sigma}\)

Substituting \(Y = 2\mu\), \(\mu\), and \(\sigma = \frac{2}{3} \mu\), we have:

\(z = \frac{2\mu - \mu}{\frac{2}{3} \mu} = \frac{\mu}{\frac{2}{3} \mu} = \frac{3}{2} = 1.5\)

Thus, \(P(Y < 2\mu) = P(z < 1.5)\).

Using standard normal distribution tables or a calculator, \(P(z < 1.5) = 0.933\).

To find the proportion of Marok’s pulse rates above 160 beats per minute, we standardize the value using the formula for the z-score:

\(z = \frac{X - \mu}{\sigma}\)

where \(X = 160\), \(\mu = 148.6\), and \(\sigma = 18.5\).

\(z = \frac{160 - 148.6}{18.5} = 0.616\)

We need to find \(P(X > 160) = P(z > 0.616)\).

Using the standard normal distribution table, \(P(z > 0.616) = 1 - P(z < 0.616) = 1 - 0.7310 = 0.269\).

Thus, the proportion of pulse rates above 160 beats per minute is 0.269.

(a) To find the probability that a randomly chosen cyclist has a time less than 74 minutes, we use the standard normal distribution. The standardization formula is:

\(Z = \frac{X - \mu}{\sigma}\)

where \(X = 74\), \(\mu = 62.3\), and \(\sigma = 8.4\). Substituting these values, we get:

\(Z = \frac{74 - 62.3}{8.4} = 1.393\)

We find \(P(Z < 1.393)\) using the standard normal distribution table, which gives us:

\(P(X < 74) = 0.918\)

(b) To find the probability that 4 randomly chosen cyclists all have times between 50 and 74 minutes, we first find the probability for one cyclist:

\(P(50 < X < 74) = P\left(\frac{50 - 62.3}{8.4} < Z < \frac{74 - 62.3}{8.4}\right)\)

\(P(-1.464 < Z < 1.393)\)

Using the standard normal distribution table, we find:

\(\Phi(1.393) = 0.9182\)

\(\Phi(-1.464) = 0.0718\)

\(P(-1.464 < Z < 1.393) = 0.9182 - 0.0718 = 0.8464\)

The probability that all 4 cyclists have times between 50 and 74 minutes is:

\((0.8464)^4 = 0.514\)

(i) To find the probability that a book is classified as large, we first standardize the height 29 cm using the normal distribution with mean 21.7 and standard deviation 6.5:

\(P(\text{large}) = 1 - \Phi\left(\frac{29 - 21.7}{6.5}\right)\)

\(= 1 - \Phi(1.123) = 1 - 0.8692 = 0.1308\)

Now, we use the binomial distribution to find the probability that fewer than 2 books are large out of 8:

\(P(0,1) = (0.8692)^8 + \binom{8}{1}(0.1308)(0.8692)^7\)

\(= 0.718\)

(ii) We need the probability of at least 1 large book to be more than 0.98. This is equivalent to:

\(1 - (0.8692)^n > 0.98\)

\((0.8692)^n < 0.02\)

Solving for the smallest integer \(n\), we find:

\(n = 28\)

Given that \(Y \sim N(\mu, \sigma^2)\) and \(2\sigma = 3\mu\), we need to find \(P(Y > 4\mu)\).

First, express \(\sigma\) in terms of \(\mu\):

\(2\sigma = 3\mu \Rightarrow \sigma = \frac{3\mu}{2}\)

Standardize the variable:

\(P(Y > 4\mu) = P\left( z > \frac{4\mu - \mu}{\sigma} \right) = P\left( z > \frac{3\mu}{\frac{3\mu}{2}} \right) = P(z > 2)\)

Using the standard normal distribution table, \(P(z > 2) = 1 - P(z \leq 2) = 1 - 0.9772 = 0.0228\).

(i) To find the number of sheep weighing between 70 kg and 72.5 kg, we first standardize the weights using the formula for the z-score:

\(z_1 = \frac{70 - 66.4}{5.6} = 0.6429\)

\(z_2 = \frac{72.5 - 66.4}{5.6} = 1.089\)

Next, we find the area under the standard normal curve between these z-scores:

\(\Phi(1.089) - \Phi(0.643) = 0.8620 - 0.7399 = 0.1221\)

Multiply this probability by the total number of sheep:

\(0.1221 \times 250 = 30.5\)

Therefore, approximately 30 or 31 sheep have a weight between 70 kg and 72.5 kg.

(ii) To find the value of \(y\), we use the symmetry of the normal distribution. The mean is 66.4 kg, and the proportion of sheep weighing less than 59.2 kg is equal to the proportion weighing more than \(y\) kg. Calculate the difference from the mean:

\(66.4 - 59.2 = 7.2\)

Thus, \(y = 66.4 + 7.2 = 73.6\) kg.

Let the random variable representing petrol consumption be denoted by \(X\), where \(X \sim N(24, 4.7^2)\).

We need to find \(P(21.6 < X < 28.7)\).

Standardize the variable using the formula:

\(z = \frac{x - \mu}{\sigma}\)

For \(x = 21.6\):

\(z_1 = \frac{21.6 - 24}{4.7} = -0.5106\)

For \(x = 28.7\):

\(z_2 = \frac{28.7 - 24}{4.7} = 1\)

Thus, we need to find \(P(-0.5106 < z < 1)\).

This is equivalent to \(\Phi(1) - \Phi(-0.5106)\).

Using standard normal distribution tables:

\(\Phi(1) = 0.8413\)

\(\Phi(-0.5106) = 1 - \Phi(0.5106) = 1 - 0.6953 = 0.3047\)

Therefore, \(P(-0.5106 < z < 1) = 0.8413 - 0.3047 = 0.537\).

First, we find the probability that a single flower pot has a base diameter between 13.6 cm and 14.8 cm. We standardize the values using the normal distribution:

\(P(13.6 < X < 14.8) = P\left( \frac{13.6 - 14}{0.52} < z < \frac{14.8 - 14}{0.52} \right)\)

\(= P(-0.7692 < z < 1.538)\)

Using the standard normal distribution table, we find:

\(= \Phi(1.538) - [1 - \Phi(0.7692)]\)

\(= 0.9380 - [1 - 0.7791]\)

\(= 0.7171\)

Now, we use the binomial distribution to find the probability that exactly 8 out of 10 flower pots have diameters in this range:

\(P(8) = (0.7171)^8 (0.2829)^2 \binom{10}{8}\)

\(= 0.252\)

To find the probability that \(X < -2.4\), we first standardize the variable using the formula for the standard normal distribution:

\(P(X < -2.4) = P\left( Z < \frac{-2.4 - 1.5}{3.2} \right)\)

Calculating the standardized value:

\(\frac{-2.4 - 1.5}{3.2} = \frac{-3.9}{3.2} = -1.219\)

Now, we find \(P(Z < -1.219)\) using the standard normal distribution table or a calculator:

\(P(Z < -1.219) = 1 - P(Z < 1.219)\)

From the standard normal distribution table, \(P(Z < 1.219) \approx 0.8886\).

Therefore, \(P(Z < -1.219) = 1 - 0.8886 = 0.1114\).

Rounding to three decimal places, the probability is \(0.111\).

The normal distribution \(X \sim N(30, 49)\) has a mean of 30 and a variance of 49, which means it is wider and shorter compared to \(Y\) and \(Z\). The curve for \(X\) should be centered at 30 and spread roughly from 10 to 50.

The normal distribution \(Y \sim N(30, 16)\) has the same mean as \(X\) but a smaller variance, making it taller and narrower. The curve for \(Y\) should also be centered at 30 but be higher and thinner than \(X\).

The normal distribution \(Z \sim N(50, 16)\) has a mean of 50 and the same variance as \(Y\). The curve for \(Z\) should be the same shape as \(Y\) but centered at 50.

To find the probability that a randomly chosen value of Y is negative, we need to standardize the variable. The standard normal variable Z is given by:

\(Z = \frac{Y - \mu}{\sigma}\)

where \(\sigma = \frac{1}{2} \mu\). Therefore, the standardization becomes:

\(Z = \frac{0 - \mu}{\frac{1}{2} \mu} = \frac{-\mu}{\frac{1}{2} \mu} = -2\)

We need to find \(P(Y < 0) = P(Z < -2)\).

Using standard normal distribution tables, \(P(Z < -2) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228\).

To find the probability that a value of X rounds to 84, we need to calculate the probability that X is between 83.5 and 84.5.

Standardize the values using the formula for the standard normal distribution:

\(Z = \frac{X - \mu}{\sigma}\)

where \(\mu = 82\) and \(\sigma = \sqrt{126}\).

Calculate the standardized values:

\(Z_1 = \frac{84.5 - 82}{\sqrt{126}} = 0.2227\)

\(Z_2 = \frac{83.5 - 82}{\sqrt{126}} = 0.1336\)

Find the probabilities using the standard normal distribution table:

\(\Phi(0.2227) = 0.5883\)

\(\Phi(0.1336) = 0.5533\)

Subtract the probabilities to find the probability that X rounds to 84:

\(P(83.5 < X < 84.5) = \Phi(0.2227) - \Phi(0.1336) = 0.5883 - 0.5533 = 0.0350\)

(i) To find the probability that the profit is between $10,000 and $12,000, we standardize these values:

\(z_1 = \frac{12 - 6.4}{5.2} = 1.077\)

\(z_2 = \frac{10 - 6.4}{5.2} = 0.692\)

Using the standard normal distribution table, find:

\(\Phi(z_1) - \Phi(z_2) = 0.8593 - 0.7556 = 0.104\)

(ii) To find the probability of a loss, we standardize the value for a loss (profit less than $0):

\(P(\text{loss}) = P\left(z < \frac{0 - 6.4}{5.2}\right) = P(z < -1.231)\)

Using the standard normal distribution table, find:

\(P(\text{loss}) = 1 - 0.8909 = 0.109\)

The probability of exactly 1 loss in 4 days is given by the binomial distribution:

\(P(1) = (0.109)^1 (0.8909)^3 \times \binom{4}{1}\)

Calculate:

\(P(1) = 0.309 \text{ or } 0.308\)

Let the random variable \(X\) represent the weight of the bags. \(X\) is normally distributed with mean \(\mu = 1.04\) kg and standard deviation \(\sigma = 0.06\) kg.

We need to find \(P(X > 1.11)\).

First, standardize the variable using the formula:

\(Z = \frac{X - \mu}{\sigma}\)

Substitute the values:

\(Z = \frac{1.11 - 1.04}{0.06} = 1.167\)

We need \(P(Z > 1.167)\).

Using standard normal distribution tables or a calculator, find \(P(Z > 1.167) = 1 - P(Z \leq 1.167)\).

From the tables, \(P(Z \leq 1.167) \approx 0.8784\).

Thus, \(P(Z > 1.167) = 1 - 0.8784 = 0.1216\).

Therefore, the probability that a randomly chosen bag weighs more than 1.11 kg is approximately \(0.122\).

To find the probability that a randomly chosen salmon is 34 cm long, we need to consider the range from 33.5 cm to 34.5 cm due to rounding to the nearest centimetre.

We standardize the values using the formula for the standard normal distribution:

\(Z = \frac{X - \mu}{\sigma}\)

For 34.5 cm: \(Z = \frac{34.5 - 32.9}{2.4} = 0.667\)

For 33.5 cm: \(Z = \frac{33.5 - 32.9}{2.4} = 0.25\)

We find the probabilities using the standard normal distribution table:

\(\Phi(0.667) = 0.7477\)

\(\Phi(0.25) = 0.5987\)

The probability that a salmon is 34 cm long is the difference between these probabilities:

\(0.7477 - 0.5987 = 0.149\)

To find the probability that \(X\) lies between 25 and 30, we standardize the values using the formula for the standard normal variable \(Z\):

\(z = \frac{x - \mu}{\sigma}\)

where \(\mu = 28.3\) and \(\sigma = \sqrt{4.5}\).

First, calculate \(z_1\) for \(x = 30\):

\(z_1 = \frac{30 - 28.3}{\sqrt{4.5}} = 0.8014\)

Next, calculate \(z_2\) for \(x = 25\):

\(z_2 = \frac{25 - 28.3}{\sqrt{4.5}} = -1.5556\)

The probability that \(X\) lies between 25 and 30 is given by:

\(\Phi(z_1) - \Phi(z_2) = \Phi(0.8014) - (1 - \Phi(-1.5556))\)

Using standard normal distribution tables or a calculator, we find:

\(\Phi(0.8014) = 0.7884\)

\(\Phi(-1.5556) = 0.0599\)

Thus, the probability is:

\(0.7884 + 0.9401 - 1 = 0.729\)

We are given that the daily minimum temperature follows a normal distribution \(N(8, 24)\), where the mean \(\mu = 8\) and the variance \(\sigma^2 = 24\). The standard deviation \(\sigma = \sqrt{24}\).

To find the probability that the temperature is between 7°C and 12°C, we standardize these values to find the corresponding \(z\)-scores.

For 12°C:

\(z_1 = \frac{12 - 8}{\sqrt{24}} = \frac{4}{\sqrt{24}} = 0.816\)

Using the standard normal distribution table, \(\Phi(0.816) = 0.7926\).

For 7°C:

\(z_2 = \frac{7 - 8}{\sqrt{24}} = \frac{-1}{\sqrt{24}} = -0.204\)

Using the standard normal distribution table, \(\Phi(-0.204) = 1 - 0.5808 = 0.4192\).

The probability that the temperature is between 7°C and 12°C is:

\(P(7 < X < 12) = \Phi(0.816) - \Phi(-0.204) = 0.7926 - 0.4192 = 0.373\)

The weights of female 18-year-old students can be modeled using a normal distribution. This is a common assumption for biological measurements, which tend to be symmetrically distributed around a mean.

For the normal distribution, we need to specify the mean and the variance. According to the mark scheme, a suitable mean is 60 kg, and a suitable variance is 90 kg².

These values are within the sensible range provided: mean between 40 kg and 80 kg, and variance between 16 kg² and 225 kg².

(i) To find the number of students expected to take longer than 20 minutes, we first standardize the variable:

\(P(X > 20) = P\left(Z > \frac{20 - 26.4}{3.7}\right) = P(Z > -1.730)\)

Using the standard normal distribution table, \(P(Z > -1.730) = 0.9582\).

For a sample of 350 students, the expected number is \(350 \times 0.9582 \approx 335 \text{ or } 336\).

(ii) A ‘very slow’ student is more than 1.645 standard deviations above the mean:

\(P(\text{very slow}) = P(Z > 1.645) = 0.05\).

We use the binomial distribution for 8 students with \(p = 0.05\):

\(P(0, 1, 2) = (0.95)^8 + 8 \times (0.05)^1 \times (0.95)^7 + \binom{8}{2} \times (0.05)^2 \times (0.95)^6\)

\(= 0.6634 + 0.2793 + 0.0515 = 0.994\).

(i) To find the probability that a pencil has a length greater than 10.9 cm, we standardize the normal distribution:

\(P(x > 10.9) = P\left(z > \frac{10.9 - 11}{0.095}\right)\)

\(= P(z > -1.0526)\)

Using the standard normal distribution table, \(P(z > -1.0526) = 0.8538\), which rounds to 0.854.

(ii) To find the probability that at least two pencils in a sample of 6 have lengths less than 10.9 cm, we first find the probability of a single pencil being less than 10.9 cm:

\(P(x < 10.9) = 1 - P(x > 10.9) = 1 - 0.8538 = 0.1462\)

We use the binomial distribution for 6 trials with probability \(p = 0.1462\):

\(P(\text{at least 2} < 10.9) = 1 - P(0, 1)\)

\(= 1 - \left(0.8538^6 + 6 \cdot 0.1462 \cdot 0.8538^5\right)\)

\(= 1 - (0.8538^6 + 6 \cdot 0.1462 \cdot 0.8538^5)\)

\(= 0.215\)

Let \(X\) be the daily minimum temperature in January in Ottawa. We have \(X \sim N(-15.1, 62.0)\).

We need to find \(P(X > 0)\).

Standardize the variable:

\(Z = \frac{X - (-15.1)}{\sqrt{62}}\)

For \(X = 0\),

\(Z = \frac{0 - (-15.1)}{\sqrt{62}} = \frac{15.1}{\sqrt{62}} \approx 1.918\)

Using the standard normal distribution table, find \(\Phi(1.918) \approx 0.9724\).

Thus, \(P(X > 0) = 1 - \Phi(1.918) = 1 - 0.9724 = 0.0276\).

First, standardize the values 1.82 and 1.92 using the formula for the z-score:

\(z_1 = \frac{1.92 - 1.9}{0.15} = 0.1333\)

\(z_2 = \frac{1.82 - 1.9}{0.15} = -0.5333\)

Next, find the probability that a single tyre has a pressure between 1.82 and 1.92 bars:

\(\text{area} = \Phi(0.1333) - \Phi(-0.5333)\)

Using standard normal distribution tables or a calculator:

\(\Phi(0.1333) = 0.5529\)

\(\Phi(-0.5333) = 1 - \Phi(0.5333) = 1 - 0.7029 = 0.2971\)

Thus, the probability for one tyre is:

\(0.5529 - 0.2971 = 0.256\)

Since the tyres are independent, the probability that all four tyres have pressures between 1.82 and 1.92 bars is:

\((0.256)^4 = 0.0043\)

(i) To find the proportion of melons classified as small, we calculate the z-score for 350 grams using the formula:

\(z = \frac{350 - 450}{120} = -0.833\)

The proportion of melons weighing less than 350 grams is given by the cumulative distribution function (CDF) for a standard normal distribution at \(z = -0.833\). From standard normal distribution tables, \(\Phi(-0.833) \approx 0.2025\). Therefore, the proportion of melons classified as small is 20.25%.

(ii) The remaining melons are divided equally between medium and large. The proportion of melons not classified as small is \(1 - 0.2025 = 0.7975\). Half of these are classified as large, so the proportion classified as large is \(0.7975 / 2 = 0.39875\).

We need to find the z-score corresponding to a cumulative probability of \(1 - 0.39875 = 0.60125\). From standard normal distribution tables, \(z \approx 0.257\).

Convert this z-score back to the original scale using:

\(x = 120 \times 0.257 + 450 = 481\)

Thus, the weight above which melons are classified as large is 481 grams.

(i) To find the proportion of people who spend between 7.8 days and 11.0 days in hospital, we use the standard normal distribution. First, calculate the z-scores:

\(z = \frac{11.0 - 7.8}{2.8} = 1.143\)

The probability that a person spends less than 11.0 days is given by:

\(P(T < 11.0) = \Phi(1.143) = 0.8735\)

Since 7.8 days is the mean, the probability of spending less than 7.8 days is 0.5. Therefore, the proportion of people who spend between 7.8 and 11.0 days is:

\(P(7.8 < T < 11.0) = 0.8735 - 0.5 = 0.3735\)

(ii) To find the probability that exactly 2 out of 3 people spend longer than 11.0 days, use the binomial distribution. The probability of one person spending more than 11.0 days is:

\(P(T > 11.0) = 1 - 0.8735 = 0.1265\)

The probability of exactly 2 out of 3 people spending more than 11.0 days is:

\((0.1265)^2 \times (0.8735) \times \binom{3}{2} = 0.0419\)

(iii) The box-and-whisker plot is not symmetric, indicating that the data is not normally distributed. Therefore, it does not agree with the model used in parts (i) and (ii).

(a) To find the number of rods expected to have a length less than 54.8 mm, we first standardize the value:

\(P(X < 54.8) = P\left(Z < \frac{54.8 - 55.6}{1.2}\right)\)

\(= P(Z < -0.6667) = 1 - 0.7477 = 0.2523\)

The expected number is then:

\(400 \times 0.2523 = 100.92\)

Rounding gives 100 or 101 rods.

(b) To find the probability that a rod's length is within half a standard deviation of the mean, we calculate:

\(P\left(-\frac{1}{2} < Z < \frac{1}{2}\right) = \Phi\left(\frac{1}{2}\right) - \Phi\left(-\frac{1}{2}\right)\)

\(= 2\Phi\left(\frac{1}{2}\right) - 1\)

\(= 2 \times 0.6915 - 1 = 0.383\)

Let the random variable \(X\) represent the height of the sunflowers, where \(X \sim N(112, 17.2^2)\).

We need to find \(P(X > 120)\).

First, standardize the variable:

\(Z = \frac{X - \mu}{\sigma} = \frac{120 - 112}{17.2} = 0.4651\)

Now, find the probability:

\(P(X > 120) = 1 - \Phi(0.4651)\)

Using standard normal distribution tables or a calculator, \(\Phi(0.4651) \approx 0.6790\).

Thus, \(P(X > 120) = 1 - 0.6790 = 0.321\)

Let the random variable representing the distance be denoted as \(X\), where \(X \sim N(35.0, 11.6^2)\).

We need to find \(P(30 < X < 40)\).

First, standardize the variable using the formula for the z-score:

\(z = \frac{x - ext{mean}}{ ext{standard deviation}}\)

For \(x = 40\):

\(z = \frac{40 - 35.0}{11.6} = 0.431\)

For \(x = 30\):

\(z = \frac{30 - 35.0}{11.6} = -0.431\)

Using the standard normal distribution table, find \(\, \Phi(0.431)\) and \(\, \Phi(-0.431)\).

\(P(30 < X < 40) = \Phi(0.431) - (1 - \Phi(0.431))\)

\(= 0.334\)

To find the proportion of cars exceeding the speed limit, we first standardize the speed limit using the z-score formula:

\(z = \frac{110 - 107.6}{13.8} = 0.174\)

Next, we find the probability that a car's speed is greater than 110 km h^-1:

\(P(X > 110) = 1 - \Phi(0.174)\)

Using standard normal distribution tables, \(\Phi(0.174) \approx 0.5691\).

Thus,

\(P(X > 110) = 1 - 0.5691 = 0.431\)

(a) To find the probability that an apple is sold to the supermarket, we need to calculate the probability that the weight of an apple is between 142 grams and 205 grams. We use the standard normal distribution:

\(P(142 < X < 205) = P\left(\frac{142 - 170}{25} < z < \frac{205 - 170}{25}\right)\)

This simplifies to:

\(P(-1.12 < z < 1.4)\)

Using the standard normal distribution table, we find:

\(\Phi(1.4) - (1 - \Phi(1.12)) = 0.9192 + 0.8686 - 1\)

Thus, the probability is:

\(0.788\)

(b) To calculate the total income, we first find the probability that an apple weighs more than 205 grams:

\(P(X > 205) = 1 - 0.9192 = 0.0808\)

The income from apples sold to the supermarket is:

\(0.788 \times 0.24 \times 20000\)

The income from apples sold to the local shop is:

\(0.0808 \times 0.30 \times 20000\)

Therefore, the total income is:

\((0.0808 \times 0.30 + 0.788 \times 0.24) \times 20000 = 4266.24\)

We need to find \(P(1.98 < X < 2.03)\) where \(X \sim N(2.02, 0.03^2)\).

First, standardize the values using the formula \(z = \frac{x - \mu}{\sigma}\).

For \(x = 1.98\), \(z = \frac{1.98 - 2.02}{0.03} = -1.333\).

For \(x = 2.03\), \(z = \frac{2.03 - 2.02}{0.03} = 0.333\).

Thus, \(P(1.98 < X < 2.03) = P(-1.333 < z < 0.333)\).

Using the standard normal distribution table, find \(\Phi(0.333)\) and \(\Phi(-1.333)\):

\(\Phi(0.333) = 0.6304\)

\(\Phi(-1.333) = 1 - \Phi(1.333) = 1 - 0.9087 = 0.0913\)

Therefore, \(P(-1.333 < z < 0.333) = \Phi(0.333) - (1 - \Phi(1.333)) = 0.6304 + 0.9087 - 1 = 0.539\).

The lengths are normally distributed with mean \(\mu = 5.2\) and standard deviation \(\sigma = 1.5\).

We need to find \(P(X < 6)\).

Standardize the variable: \(Z = \frac{X - \mu}{\sigma}\).

Substitute the values: \(Z = \frac{6 - 5.2}{1.5} = 0.5333\).

Find \(P(Z < 0.5333)\) using standard normal distribution tables or a calculator.

The probability is approximately 0.703.