Past Exam Questions

(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, \(P(z < 0.5) = 0.692\).