Exam question practice Problem set B Worked solutions

The mass, X kilograms, of a pumpkin is normally distributed with mean 3.4 and standard deviation 0.6. Find the probability that

1. (a) one pumpkin has a mass greater than 4 kg;

   • \(X \sim N(3.4, 0.6^2)\)
   • From the calculator: normal cdf, lower bound 4, upper bound 1E99, mean 3.4, standard deviation 0.6: 0.159
   • P(X > 4) = 0.159

2. (b) two pumpkins chosen at random both have a mass greater than 4 kg.

   • Two independent items, both above 4: multiply.
   • P(both) = 0.1587² = 0.0252

The mass, X kilograms, of a pumpkin is normally distributed with mean 3.4 and standard deviation 0.6. Two pumpkins are chosen at random. Find the probability that their total mass is greater than 7.5 kg.

• \(T = X_1 + X_2\): E(T) = 3.4 + 3.4 = 6.8
• Var(T) = 0.6² + 0.6² = 0.72, so T ~ N(6.8, 0.72)
• From the calculator: normal cdf, lower bound 7.5, upper bound 1E99, mean 6.8, standard deviation 0.8485: 0.205
• P(T > 7.5) = 0.205

A grower uses a new fertiliser, which increases the mass of every pumpkin by 5%. Before the change the mass has mean 3.4 kg and standard deviation 0.6 kg.

1. (a) Find the new mean.

   • The new mass is 1.05 times the old mass.
   • New mean = 1.05 × 3.4 = 3.57

2. (b) Find the new standard deviation.

   • The standard deviation is multiplied by 1.05 too.
   • New standard deviation = 1.05 × 0.6 = 0.63

A grower thinks that a new fertiliser has increased the mean mass of a pumpkin. The old mean was 3.4 kg and the standard deviation is 0.6 kg. He tests a random sample of 20 pumpkins at the 5% significance level.

1. (a) State the null and alternative hypotheses.

   • H1 is one-tailed (an increase), and the value 3.4 appears in both.
   • H0: μ = 3.4
   • H1: μ > 3.4

2. (b) Find the critical region for the sample mean. Give the critical value to 3 significant figures.

   • H0: μ = 3.4, H1: μ > 3.4
   • Under H0, \(\bar X \sim N(3.4, 0.6^2/20)\), standard deviation 0.6/√20 = 0.134
   • From the calculator: inverse normal, area 0.95, mean 3.4, standard deviation 0.134: 3.62
   • Critical region: \(\bar X > 3.62\)

The new fertiliser in fact increases the mass of every pumpkin by 5%. A test of H0: μ = 3.4 against H1: μ > 3.4 uses a sample of 20 pumpkins, a standard deviation of 0.6 kg and a 5% significance level. Before the change the mean mass is 3.4 kg.

1. (a) Find the probability of a Type II error. Use the unrounded critical value.

   • \(W = 1.05X\)
   • E(W) = 1.05 × 3.4 = 3.57
   • SD(W) = 1.05 × 0.6 = 0.63
   • True distribution: \(\bar X \sim N(3.57, 0.63^2/20)\), standard deviation 0.63/√20 = 0.141
   • Type II error: \(\bar X < 3.62068\)
   • From the calculator: normal cdf, lower bound −1E99, upper bound 3.62068 (the unrounded critical value), mean 3.57, standard deviation 0.141: 0.640
   • P(Type II error) = 0.640

The mass, X kilograms, of a pumpkin is normally distributed with mean 3.4 and standard deviation 0.6. A grower uses a new fertiliser and thinks the mean mass has increased. He tests this at the 5% significance level, using a random sample of 20 pumpkins.

1. (a) Find the probability that a pumpkin has a mass greater than 4.0 kg.

   • \(X \sim N(3.4, 0.6^2)\)
   • From the calculator: normal cdf, lower bound 4, upper bound 1E99, mean 3.4, standard deviation 0.6: 0.159
   • P(X > 4) = 0.159

2. (b) Two pumpkins are chosen at random. Find the probability that their total mass is greater than 7.5 kg.

   • \(T = X_1 + X_2\): E(T) = 3.4 + 3.4 = 6.8
   • Var(T) = 0.6² + 0.6² = 0.72, so T ~ N(6.8, 0.72)
   • From the calculator: normal cdf, lower bound 7.5, upper bound 1E99, mean 6.8, standard deviation 0.8485: 0.205
   • P(T > 7.5) = 0.205

3. (c) State the null and alternative hypotheses.

   • H0: μ = 3.4
   • H1: μ > 3.4

4. (d) Find the critical region for the sample mean.

   • Under H0, \(\bar X \sim N(3.4, 0.6^2/20)\), standard deviation 0.6/√20 = 0.134
   • From the calculator: inverse normal, area 0.95, mean 3.4, standard deviation 0.134: 3.62
   • Critical region: \(\bar X > 3.62\)

5. (e) The fertiliser in fact makes every pumpkin 5% heavier. Find the new mean and standard deviation of the mass of a pumpkin.

   • \(W = 1.05X\)
   • E(W) = 1.05 × 3.4 = 3.57
   • SD(W) = 1.05 × 0.6 = 0.63

6. (f) Find the probability of a Type II error. Use the unrounded critical value.

   • True distribution: \(\bar X \sim N(3.57, 0.63^2/20)\), standard deviation 0.63/√20 = 0.141
   • Type II error: \(\bar X < 3.62068\)
   • From the calculator: normal cdf, lower bound −1E99, upper bound 3.62068 (the unrounded critical value), mean 3.57, standard deviation 0.141: 0.640
   • P(Type II error) = 0.640