Exam question practice Problem set A One question at a time

The mass, X grams, of a baguette is normally distributed with mean 260 and standard deviation 10. Find the probability that

1. (a) one baguette has a mass greater than 272 g;

   • \(X \sim N(260, 10^2)\)
   • From the calculator: normal cdf, lower bound 272, upper bound 1E99, mean 260, standard deviation 10: 0.115
   • P(X > 272) = 0.115

2. (b) two baguettes chosen at random both have a mass greater than 272 g.

   • Two independent items, both above 272: multiply.
   • P(both) = 0.1151² = 0.0132

The mass, X grams, of a baguette is normally distributed with mean 260 and standard deviation 10. Two baguettes are chosen at random. Find the probability that their total mass is greater than 540 g.

• \(T = X_1 + X_2\): E(T) = 260 + 260 = 520
• Var(T) = 10² + 10² = 200, so T ~ N(520, 200)
• From the calculator: normal cdf, lower bound 540, upper bound 1E99, mean 520, standard deviation 14.14: 0.0786
• P(T > 540) = 0.0786

A bakery changes its flour, which increases the mass of every baguette by 2%. Before the change the mass has mean 260 g and standard deviation 10 g.

1. (a) Find the new mean.

   • The new mass is 1.02 times the old mass.
   • New mean = 1.02 × 260 = 265.2

2. (b) Find the new standard deviation.

   • The standard deviation is multiplied by 1.02 too.
   • New standard deviation = 1.02 × 10 = 10.2

A baker thinks that a new recipe has increased the mean mass of a baguette. The old mean was 260 g and the standard deviation is 10 g. She tests a random sample of 16 baguettes at the 5% significance level.

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

   • H1 is one-tailed (an increase), and the value 260 appears in both.
   • H0: μ = 260
   • H1: μ > 260

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

   • H0: μ = 260, H1: μ > 260
   • Under H0, \(\bar X \sim N(260, 10^2/16)\), standard deviation 10/√16 = 2.50
   • From the calculator: inverse normal, area 0.95, mean 260, standard deviation 2.50: 264
   • Critical region: \(\bar X > 264\)

The new recipe in fact increases the mass of every baguette by 2%. A test of H0: μ = 260 against H1: μ > 260 uses a sample of 16 baguettes, a standard deviation of 10 g and a 5% significance level. Before the change the mean mass is 260 g.

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

   • \(W = 1.02X\)
   • E(W) = 1.02 × 260 = 265.2
   • SD(W) = 1.02 × 10 = 10.2
   • True distribution: \(\bar X \sim N(265.2, 10.2^2/16)\), standard deviation 10.2/√16 = 2.55
   • Type II error: \(\bar X < 264.112\)
   • From the calculator: normal cdf, lower bound −1E99, upper bound 264.112 (the unrounded critical value), mean 265.2, standard deviation 2.55: 0.335
   • P(Type II error) = 0.335

The mass, X grams, of a baguette is normally distributed with mean 260 and standard deviation 10. A baker changes the recipe and thinks the mean mass has increased. She tests this at the 5% significance level, using a random sample of 16 baguettes.

1. (a) Find the probability that a baguette has a mass greater than 272 g.

   • \(X \sim N(260, 10^2)\)
   • From the calculator: normal cdf, lower bound 272, upper bound 1E99, mean 260, standard deviation 10: 0.115
   • P(X > 272) = 0.115

2. (b) Two baguettes are chosen at random. Find the probability that their total mass is greater than 540 g.

   • \(T = X_1 + X_2\): E(T) = 260 + 260 = 520
   • Var(T) = 10² + 10² = 200, so T ~ N(520, 200)
   • From the calculator: normal cdf, lower bound 540, upper bound 1E99, mean 520, standard deviation 14.14: 0.0786
   • P(T > 540) = 0.0786

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

   • H0: μ = 260
   • H1: μ > 260

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

   • Under H0, \(\bar X \sim N(260, 10^2/16)\), standard deviation 10/√16 = 2.50
   • From the calculator: inverse normal, area 0.95, mean 260, standard deviation 2.50: 264
   • Critical region: \(\bar X > 264\)

5. (e) The new recipe in fact makes every baguette 2% heavier. Find the new mean and standard deviation of the mass of a baguette.

   • \(W = 1.02X\)
   • E(W) = 1.02 × 260 = 265.2
   • SD(W) = 1.02 × 10 = 10.2

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

   • True distribution: \(\bar X \sim N(265.2, 10.2^2/16)\), standard deviation 10.2/√16 = 2.55
   • Type II error: \(\bar X < 264.112\)
   • From the calculator: normal cdf, lower bound −1E99, upper bound 264.112 (the unrounded critical value), mean 265.2, standard deviation 2.55: 0.335
   • P(Type II error) = 0.335