Past Exam Questions

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:

\(0.7884 + 0.9401 - 1 = 0.729\)