Past Exam Questions

The number of times the gymnast performs the routine correctly follows a binomial distribution with parameters \(n = 50\) and \(p = 0.65\).

We approximate this binomial distribution with a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) of the binomial distribution are given by:

\(\mu = n \times p = 50 \times 0.65 = 32.5\)

\(\sigma^2 = n \times p \times (1-p) = 50 \times 0.65 \times 0.35 = 11.375\)

We use a continuity correction to find the probability of fewer than 29 occasions:

\(P(X < 29) \approx P\left(Z < \frac{28.5 - 32.5}{\sqrt{11.375}}\right)\)

\(= P\left(Z < \frac{-4}{\sqrt{11.375}}\right)\)

\(= P(Z < -1.186)\)

Using the standard normal distribution table, \(P(Z < -1.186) = 1 - P(Z < 1.186) = 1 - 0.8822 = 0.118\).

Let \(X\) be the number of years out of 56 where New Year's Day is on a Saturday. \(X\) follows a binomial distribution with parameters \(n = 56\) and \(p = \frac{1}{7}\).

We approximate this binomial distribution with a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) are given by:

\(\mu = 56 \times \frac{1}{7} = 8\)

\(\sigma^2 = 56 \times \frac{1}{7} \times \frac{6}{7} = 6.857\)

We need to find \(P(X > 7)\). Using a continuity correction, this becomes \(P(X > 7.5)\).

Standardizing, we have:

\(P(X > 7.5) = 1 - \Phi \left( \frac{7.5 - 8}{\sqrt{6.857}} \right)\)

\(= 1 - \Phi(-0.1909)\)

\(= \Phi(0.1909)\)

Using standard normal distribution tables, \(\Phi(0.1909) \approx 0.576\).

Let the number of small bands be a random variable, denoted by \(X\). Given that 60% of the bands are small, \(X\) follows a binomial distribution \(B(n=150, p=0.6)\).

We approximate this binomial distribution with a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) of the binomial distribution are given by:

\(\mu = np = 150 \times 0.6 = 90\)

\(\sigma^2 = np(1-p) = 150 \times 0.6 \times 0.4 = 36\)

Thus, the standard deviation \(\sigma\) is \(\sqrt{36} = 6\).

We use a normal approximation to find \(P(88 < X < 97)\). Applying continuity correction, this becomes \(P(87.5 < X < 97.5)\).

Standardizing, we calculate:

\(P(87.5 < X < 97.5) = \Phi\left(\frac{97.5 - 90}{6}\right) - \Phi\left(\frac{87.5 - 90}{6}\right)\)

\(= \Phi(1.25) - \Phi(-0.4166)\)

Using standard normal distribution tables, \(\Phi(1.25) = 0.8944\) and \(\Phi(-0.4166) = 1 - 0.6616 = 0.3384\).

Thus, the probability is:

\(0.8944 - 0.3384 = 0.556\)

Let the random variable \(X\) represent the number of adults wearing a watch on their left wrist. Given that 76% wear a watch on their left wrist, \(X\) follows a binomial distribution \(B(n=200, p=0.76)\).

We approximate this binomial distribution with a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) are calculated as follows:

\(\mu = np = 200 \times 0.76 = 152\)

\(\sigma^2 = np(1-p) = 200 \times 0.76 \times 0.24 = 36.48\)

We use a continuity correction to find \(P(X > 155)\), which is approximated by \(P(X > 155.5)\) in the normal distribution.

Standardizing, we have:

\(Z = \frac{155.5 - 152}{\sqrt{36.48}} = \frac{3.5}{6.036} \approx 0.5795\)

Thus, \(P(X > 155) = 1 - \Phi(0.5795) \approx 1 - 0.7188 = 0.281\).

(i) We know that 10% of the bulbs last longer than 5130 hours. This corresponds to a z-score of 1.282 in a standard normal distribution. Using the formula for the z-score:

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

where \(X = 5130\), \(\sigma = 40.6\), and \(z = 1.282\), we solve for \(\mu\):

\(1.282 = \frac{5130 - \mu}{40.6}\)

\(\mu = 5130 - 1.282 \times 40.6\)

\(\mu = 5080\) (rounded to the nearest hour)

(ii) To find the probability that a light bulb fails to last for 5000 hours, we calculate the z-score:

\(z = \frac{5000 - 5078}{40.6} = -1.921\)

The probability \(P(X < 5000) = \Phi(-1.921) = 0.0273\) or 2.73%.

(iii) For 600 light bulbs, the mean number lasting longer than 5130 hours is \(\mu = 60\) and variance \(\sigma^2 = 54\). We approximate using a normal distribution:

\(P(\text{fewer than 65}) = \Phi\left(\frac{64.5 - 60}{\sqrt{54}}\right)\)

\(= \Phi(0.6123)\)

\(= 0.730\) or 0.73

Let the random variable \(X\) represent the number of overweight people in the sample of 400. Given that 2 out of 5 people are overweight, the probability \(p\) of a person being overweight is \(p = \frac{2}{5} = 0.4\).

The expected number of overweight people \(\mu\) is given by \(\mu = np = 400 \times 0.4 = 160\).

The variance \(\sigma^2\) is given by \(\sigma^2 = np(1-p) = 400 \times 0.4 \times 0.6 = 96\).

We use a normal approximation to the binomial distribution. The standard deviation \(\sigma\) is \(\sqrt{96} \approx 9.798\).

We apply continuity correction for the normal approximation: \(P(X < 165) \approx P(X \leq 164.5)\).

Standardizing, we find:

\(Z = \frac{164.5 - 160}{\sqrt{96}} = \frac{4.5}{9.798} \approx 0.4593\)

Using standard normal distribution tables, \(\Phi(0.4593) \approx 0.677\).

Thus, the probability that fewer than 165 people in the sample are overweight is approximately 0.677.

Let the random variable \(X\) represent the number of cars made by Ford in a sample of 50 cars. \(X\) follows a binomial distribution with parameters \(n = 50\) and \(p = 0.28\).

To use a normal approximation, we calculate the mean \(\mu\) and variance \(\sigma^2\) of the distribution:

\(\mu = 50 \times 0.28 = 14\)

\(\sigma^2 = 50 \times 0.28 \times 0.72 = 10.08\)

We apply a continuity correction to find \(P(X > 18)\), which is approximated by \(P(X > 18.5)\).

Standardize the variable:

\(Z = \frac{18.5 - 14}{\sqrt{10.08}} = 1.417\)

Using the standard normal distribution table, find \(\Phi(1.417) \approx 0.9218\) or \(0.9217\).

Thus, \(P(X > 18) = 1 - \Phi(1.417) = 1 - 0.9218 = 0.0782\) or \(1 - 0.9217 = 0.0783\).

Let the random variable \(X\) represent the number of students with blue eyes in a sample of 80 students. \(X\) follows a binomial distribution with parameters \(n = 80\) and \(p = 0.32\).

To approximate, we use a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) are given by:

\(\mu = np = 80 \times 0.32 = 25.6\)

\(\sigma^2 = np(1-p) = 80 \times 0.32 \times 0.68 = 17.408\)

Using the normal approximation, we apply continuity correction for \(P(X < 20)\):

\(P(X < 20) = P\left(Z < \frac{19.5 - 25.6}{\sqrt{17.408}}\right) = P(Z < -1.462)\)

Using standard normal distribution tables or a calculator, find:

\(P(Z < -1.462) = 1 - \Phi(1.462) = 1 - 0.9282 = 0.0718\)

Thus, the probability that fewer than 20 students have blue eyes is approximately \(0.0718 \leq p < 0.0719\).

Let the random variable \(X\) represent the number of damaged tapes in a sample of 1600 tapes. Given that 1 in 5 tapes are damaged, the probability of a tape being damaged is \(p = 0.2\).

The expected number of damaged tapes is \(\mu = np = 1600 \times 0.2 = 320\).

The variance is \(\sigma^2 = np(1-p) = 1600 \times 0.2 \times 0.8 = 256\).

We approximate the binomial distribution with a normal distribution: \(X \sim N(320, 256)\).

We need to find \(P(X \geq 290)\). Using continuity correction, this is \(P(X > 289.5)\).

Standardize: \(Z = \frac{289.5 - 320}{\sqrt{256}} = \frac{-30.5}{16} = -1.906\).

Thus, \(P(X \geq 290) = 1 - \Phi(-1.906) = \Phi(1.906)\).

Using standard normal distribution tables, \(\Phi(1.906) = 0.972\).

(i) The number of eggs laid by the hens follows a binomial distribution with parameters \(n = 30\) and \(p = 0.7\). We need to find \(P(X = 24)\).

Using the binomial probability formula:

\(P(X = 24) = \binom{30}{24} (0.7)^{24} (0.3)^{6}\)

\(= (0.7)^{24} \times (0.3)^{6} \times \binom{30}{24}\)

\(= 0.0829\)

(ii) To approximate the probability that fewer than 20 eggs are laid, we use a normal approximation to the binomial distribution.

The mean \(\mu\) and variance \(\sigma^2\) of the binomial distribution are given by:

\(\mu = np = 30 \times 0.7 = 21\)

\(\sigma^2 = np(1-p) = 30 \times 0.7 \times 0.3 = 6.3\)

Using the normal approximation, we standardize the variable:

\(P(X < 20) \approx P\left(Z < \frac{19.5 - 21}{\sqrt{6.3}}\right)\)

\(= P(Z < -0.5976)\)

Using standard normal distribution tables:

\(P(Z < -0.5976) = 1 - 0.7251 = 0.275\)

(i) Let the probability of choosing a chocolate biscuit be \(p = \frac{1}{4}\) and a cream biscuit be \(1 - p = \frac{3}{4}\). The probability of having 5 chocolate and 5 cream biscuits in a box of 10 is given by:

\(P(\text{equal}) = \left(0.25\right)^5 \times \left(0.75\right)^5 \times \binom{10}{5} = 0.0584\)

(ii) The probability that exactly 1 out of 8 boxes contains equal numbers of cream and chocolate biscuits is:

\((0.0584)^1 \times (0.9416)^7 \times \binom{8}{1} = 0.307\)

(iii) For a large box of 120 biscuits, use the normal approximation. Let \(X\) be the number of chocolate biscuits. Then \(\mu = 120 \times 0.25 = 30\) and \(\sigma^2 = 30 \times 0.75 = 22.5\).

Using continuity correction, find \(P(X < 35)\):

\(P(X < 35) = \Phi\left(\frac{34.5 - 30}{\sqrt{22.5}}\right) = \Phi(0.949)\)

From standard normal tables, \(\Phi(0.949) = 0.829\).

Let the random variable \(X\) represent the number of plants producing pink flowers in a box of 100. \(X\) follows a binomial distribution with parameters \(n = 100\) and \(p = 0.3\).

To approximate this binomial distribution, we use a normal distribution with mean \(\mu\) and variance \(\sigma^2\).

Calculate the mean: \(\mu = np = 100 \times 0.3 = 30\).

Calculate the variance: \(\sigma^2 = np(1-p) = 100 \times 0.3 \times 0.7 = 21\).

Thus, \(X \sim N(30, 21)\).

We need to find \(P(X < 35)\). Using a continuity correction, this becomes \(P(X < 34.5)\).

Standardize the variable: \(Z = \frac{34.5 - 30}{\sqrt{21}} = \frac{4.5}{\sqrt{21}} \approx 0.9820\).

Using the standard normal distribution table, \(\Phi(0.9820) \approx 0.837\).

Therefore, the probability that a box contains fewer than 35 plants which produce pink flowers is \(0.837\).

Let the random variable \(X\) represent the number of callers connected immediately. \(X\) follows a binomial distribution with parameters \(n = 80\) and \(p = 0.9\).

To approximate, use a normal distribution: \(X \sim N(\mu, \sigma^2)\).

Calculate the mean: \(\mu = np = 80 \times 0.9 = 72\).

Calculate the variance: \(\sigma^2 = npq = 80 \times 0.9 \times 0.1 = 7.2\).

Use continuity correction for \(P(X > 69)\):

\(P(X > 69) = P(Z > \frac{69.5 - 72}{\sqrt{7.2}})\).

Calculate the standard score:

\(Z = \frac{69.5 - 72}{\sqrt{7.2}} = -0.9317\).

Find the probability using the standard normal distribution:

\(P(Z > -0.9317) = \Phi(0.9317) \approx 0.824\).

For part (c), justify the approximation:

Both \(np = 72\) and \(nq = 8\) are greater than 5, so the normal approximation is valid.

Let the random variable \(X\) represent the number of residents who rate the service as 'good'. \(X\) follows a binomial distribution with parameters \(n = 125\) and \(p = 0.52\).

To approximate this binomial distribution, we use a normal distribution with mean \(\mu = np = 0.52 \times 125 = 65\) and variance \(\sigma^2 = np(1-p) = 0.52 \times 0.48 \times 125 = 31.2\).

Thus, \(X \sim N(65, 31.2)\).

We need to find \(P(X > 72)\). Using a continuity correction, this is approximated by \(P(X > 72.5)\).

Standardize the variable: \(Z = \frac{X - \mu}{\sigma} = \frac{72.5 - 65}{\sqrt{31.2}} \approx 1.343\).

Thus, \(P(X > 72.5) = P(Z > 1.343) = 1 - P(Z \leq 1.343)\).

Using standard normal distribution tables, \(P(Z \leq 1.343) \approx 0.9104\).

Therefore, \(P(Z > 1.343) = 1 - 0.9104 = 0.0896\).

Hence, the probability that more than 72 residents rate the service as 'good' is approximately between 0.0896 and 0.0897.

(a) Use the formula for distance traveled in the nth second:

\(s = u + \frac{1}{2}a(2n-1)\).

For the 2nd second:

\(52 = u + \frac{1}{2}a(3)\).

For the 4th second:

\(64 = u + \frac{1}{2}a(7)\).

These simplify to:

\(52 = u + 1.5a\)

\(64 = u + 3.5a\).

Subtract the first equation from the second:

\(12 = 2a\)

\(a = 6 \text{ m/s}^2\).

Substitute \(a = 6\) into \(52 = u + 1.5a\):

\(52 = u + 9\)

\(u = 43 \text{ m/s}\).

(b) Use the formula for total distance traveled:

\(s = ut + \frac{1}{2}at^2\).

Substitute \(u = 43\), \(a = 6\), and \(t = 10\):

\(s = 43 \times 10 + \frac{1}{2} \times 6 \times 10^2\)

\(s = 430 + 300\)

\(s = 730 \text{ m}\).

To solve this problem, we can use the equations of motion and the concept of areas under a velocity-time graph.

Let the constant speed be denoted by \(V\).

1. During the first 12 seconds, the bus accelerates from rest to speed \(V\). The distance covered is given by:

\(\text{Distance} = \frac{1}{2} \times 12 \times V = 6V\)

2. During the next 30 seconds, the bus travels at constant speed \(V\). The distance covered is:

\(\text{Distance} = 30V\)

3. During the final 6 seconds, the bus decelerates to rest. The distance covered is:

\(\text{Distance} = \frac{1}{2} \times 6 \times V = 3V\)

The total distance is the sum of these distances:

\(6V + 30V + 3V = 585\)

\(39V = 585\)

\(V = \frac{585}{39} = 15 \text{ m/s}\)

Thus, the constant speed of the bus is 15 m/s.

4. To find the magnitude of the deceleration, we use the equation:

\(V = u + at\)

where \(u = 15 \text{ m/s}\), \(V = 0 \text{ m/s}\), and \(t = 6 \text{ s}\).

\(0 = 15 + a \times 6\)

\(a = -\frac{15}{6} = -2.5 \text{ m/s}^2\)

The magnitude of the deceleration is 2.5 m/s².

(a) Use the constant acceleration equations to find expressions for \(s_{AB}\) and \(s_{BC}\):

\(s_{AB} = 2 \times 4.5 - \frac{1}{2} a \times 2^2\)

\(s_{BC} = 2 \times 4.5 + \frac{1}{2} a \times 2^2\)

Given \(s_{AB} = \frac{4}{5} s_{BC}\), substitute:

\([2 \times 4.5 - \frac{1}{2} a \times 2^2] = \frac{4}{5} [2 \times 4.5 + \frac{1}{2} a \times 2^2]\)

Solve for \(a\):

\(a = 0.5 \text{ m s}^{-2}\)

(b) Find \(AC\):

\(s_{AB} = 8 \text{ m}\) and \(s_{BC} = 10 \text{ m}\)

\(s_{AC} = s_{AB} + s_{BC} = 8 + 10 = 18 \text{ m}\)

(i) Using the equation of motion: \(s = ut + \frac{1}{2}at^2\).

For PQ: \(s_{PQ} = 20 \times 10 - 0.5 \times a \times 10^2\).

For QR: \(s_{QR} = 20 \times 10 + 0.5 \times a \times 10^2\).

Given \(s_{QR} = 1.5 \times s_{PQ}\), we have:

\(1.5(200 - 50a) = 200 + 50a\).

Simplifying gives \(100 = 125a\) leading to \(a = 0.8 \text{ m s}^{-2}\).

(ii) For distance QS, use \(s = ut + \frac{1}{2}at^2\).

\(s_{QS} = 20 \times 20 + \frac{1}{2} \times 0.8 \times 20^2\).

\(s_{QS} = 400 + 160 = 560 \text{ m}\).

Average speed between Q and S is \(\frac{560}{20} = 28 \text{ m s}^{-1}\).

We use the equation for motion under constant acceleration: \(s = ut + \frac{1}{2}at^2\).

For the first 80 m in 5 s:

\(80 = 5u + \frac{1}{2}a \times 5^2\)

\(80 = 5u + 12.5a\)

For the first 160 m in 8 s:

\(160 = 8u + \frac{1}{2}a \times 8^2\)

\(160 = 8u + 32a\)

We now have two simultaneous equations:

1. \(5u + 12.5a = 80\)
2. \(8u + 32a = 160\)

Solving these equations, we first multiply the first equation by 8 and the second by 5 to eliminate \(u\):

\(40u + 100a = 640\)

\(40u + 160a = 800\)

Subtract the first from the second:

\(60a = 160\)

\(a = \frac{8}{3}\)

Substitute \(a = \frac{8}{3}\) back into the first equation:

\(5u + 12.5 \times \frac{8}{3} = 80\)

\(5u + \frac{100}{3} = 80\)

\(5u = 80 - \frac{100}{3}\)

\(5u = \frac{240}{3} - \frac{100}{3}\)

\(5u = \frac{140}{3}\)

\(u = \frac{28}{3}\)

Thus, \(a = \frac{8}{3} \text{ m s}^{-2}\) and \(u = \frac{28}{3} \text{ m s}^{-1}\).

(i) Using the equation of motion:

\(s = ut + \frac{1}{2}at^2\)

For AB:

\(100 = 4u + 8a\)

For BC:

\(148 = 4v + 8a\)

Using the equation \(v = u + at\) for AB to BC:

\(v = u + 4a\)

Substitute \(v = u + 4a\) into the equation for BC:

\(148 = 4(u + 4a) + 8a\)

Simplifying gives:

\(148 = 4u + 24a\)

Now solve the system of equations:

\(100 = 4u + 8a\)

\(148 = 4u + 24a\)

Subtract the first from the second:

\(48 = 16a\)

Thus, \(a = 3\) m s^-2.

Substitute \(a = 3\) into \(100 = 4u + 8a\):

\(100 = 4u + 24\)

\(76 = 4u\)

\(u = 19\) m s^-1.

(ii) Using the equation of motion:

\(v^2 = u^2 + 2as\)

For AD:

\(61^2 = 19^2 + 2 \times 3 \times s\)

\(3721 = 361 + 6s\)

\(3360 = 6s\)

\(s = 560\)

Distance CD is:

\(CD = 560 - 248 = 312\) m.