Past Exam Questions

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.

Given that the weights are normally distributed with mean \(\mu = 0.75\) kg and standard deviation \(\sigma\) kg, and 68% of the weights are less than 0.9 kg, we can use the standard normal distribution to find \(\sigma\).

The probability statement is:

\(P(X < 0.9) = P\left( Z < \frac{0.9 - 0.75}{\sigma} \right) = 0.68\)

From standard normal distribution tables, \(P(Z < 0.468) \approx 0.68\).

Thus, \(\frac{0.9 - 0.75}{\sigma} = 0.468\).

Solving for \(\sigma\):

\(\sigma = \frac{0.15}{0.468} \approx 0.3205\)

Therefore, the value of \(\sigma\) is approximately between 0.3205 and 0.3215.

(b) We use the standard normal distribution to find the mean \(\mu\) and standard deviation \(\sigma\). Given:

\(z_1 = \frac{3 - \mu}{\sigma} = -1.329\)

\(z_2 = \frac{8 - \mu}{\sigma} = 0.878\)

Solving these equations simultaneously, we find:

\(\sigma = 2.27, \mu = 6.01\)

(c) To find the number of leaves more than 1 standard deviation from the mean, we calculate:

\(P(Z < -1) + P(Z > 1) = 2 \times \Phi(1) - \Phi(-1)\)

\(= 2 - 2 \times 0.8413 = 0.3174\)

Number of leaves: \(2000 \times 0.3174 = 634.8\), so approximately 634 or 635 leaves.

(i) Given that 5% of the fish are longer than 50 cm, we use the z-score for 95% which is 1.645. The equation is:

\(1.645 = \frac{50 - 38}{\sigma}\)

Solving for \(\sigma\), we get:

\(\sigma = \frac{50 - 38}{1.645} = 7.29\)

(ii) To find the proportion of fish shorter than 30 cm, we standardize:

\(z = \frac{30 - 38}{7.29} = -1.097\)

Using the standard normal distribution table, \(P(z < -1.097) = 1 - \Phi(1.097)\)

\(= 1 - 0.8637 = 0.136\)

(iii) The probability that a fish is longer than 50 cm is 0.05. For 9 fish, the probability that none are longer than 50 cm is \((0.95)^9\).

The probability that at least one is longer than 50 cm is:

\(1 - (0.95)^9 = 0.370\)

Given that the tyre pressures follow a normal distribution with mean \(\mu = 1.9\) bars and standard deviation \(\sigma = 0.15\) bars, we need to find the value of \(b\) such that 80% of the data lies within \(1.9 - b\) and \(1.9 + b\).

Since 80% of the data is within these limits, we have \(P(1.9 - b < X < 1.9 + b) = 0.8\).

This corresponds to a cumulative probability of 0.8, which means the z-scores are \(\pm 1.282\) (or approximately \(\pm 1.28\)).

Using the z-score formula: \(z = \frac{b}{0.15}\), we have:

\(\pm 1.282 = \frac{b}{0.15}\)

Solving for \(b\), we get:

\(b = 1.282 \times 0.15 = 0.1923\)

Thus, the safety limits are:

\(1.9 - 0.1923 = 1.7077 \approx 1.71\)

\(1.9 + 0.1923 = 2.0923 \approx 2.09\)

Therefore, the limits are between 1.71 and 2.09 bars.

(i) We know that the probability of the lunch break lasting more than 52 minutes is \(\frac{1}{4} = 0.25\). This corresponds to a standard normal value \(z\) such that \(P(Z > z) = 0.25\). From standard normal distribution tables, \(z \approx 0.674\).

Using the formula for the standard normal variable, \(z = \frac{52 - \mu}{5}\), we have:

\(\frac{52 - \mu}{5} = 0.674\)

Solving for \(\mu\):

\(52 - \mu = 5 \times 0.674\)

\(52 - \mu = 3.37\)

\(\mu = 52 - 3.37 = 48.63\)

Rounding to one decimal place, \(\mu = 48.6\).

(ii) We need to find the probability that the lunch break lasts between 40 and 46 minutes. First, standardize these values:

\(z_1 = \frac{40 - 48.63}{5} = -1.726\)

\(z_2 = \frac{46 - 48.63}{5} = -0.526\)

Using the standard normal distribution table, find the probabilities:

\(P(Z < -1.726) \approx 0.0422\)

\(P(Z < -0.526) \approx 0.2995\)

The probability that the lunch break lasts between 40 and 46 minutes is:

\(P(-1.726 < Z < -0.526) = 0.2995 - 0.0422 = 0.2573\)

The probability that this happens on every one of the next four days is:

\((0.2573)^4 = 0.00438\)