Mean, variance and binomial practice Problem set B One question at a time

\(X \sim B(45, 0.2)\). Find

1. (a) the mean of X

   • E(X) = np = 45 × 0.2 = 9

2. (b) the variance of X

   • Var(X) = np(1 − p) = 45 × 0.2 × 0.8 = 7.2

3. (c) the standard deviation of X

   • SD = √7.2 = 2.68

\(Y \sim B(26, 0.4)\). Find

1. (a) \(P(Y = 11)\)

   • From the calculator: binomial pdf with n = 26, p = 0.4, x = 11
   • P(Y = 11) = 0.152

2. (b) \(P(Y > 12)\)

   • P(Y > 12) = 1 − P(Y ≤ 12)
   • From the calculator: binomial cdf with n = 26, p = 0.4, lower = 0, upper = 12
   • P(Y ≤ 12) = 0.8007
   • P(Y > 12) = 1 − 0.8007 = 0.199

3. (c) \(P(8 \le Y < 12)\)

   • P(8 ≤ Y < 12) = P(8 ≤ Y ≤ 11)
   • From the calculator: binomial cdf with n = 26, p = 0.4, lower = 8, upper = 11
   • P(8 ≤ Y ≤ 11) = 0.552

Each of 6 pupils in a club independently walks to school with probability 0.65. X is the number of pupils who walk to school.

1. (a) Complete the table of the probability distribution of X. Give each value to 3 significant figures.

   • X ~ B(6, 0.65)
   • From the calculator: binomial pdf with n = 6, p = 0.65, x left blank gives P(X = 0) to P(X = 6)
   • x = 0 to 6: 0.00184, 0.0205, 0.0951, 0.235, 0.328, 0.244, 0.0754
   • The total is 1, apart from rounding

2. (b) Write down the most likely value of X.

   • The largest probability is 0.328, at x = 4

A carton holds 9 yoghurt pots. Each pot is damaged with probability 0.03, independently. A cafe orders 14 cartons and returns any carton that has at least one damaged pot.

1. (a) Find the probability that a carton has at least one damaged pot.

   • X = number of damaged pots in a carton of 9 yoghurt pots. X ~ B(9, 0.03)
   • P(X ≥ 1) = 1 − P(X = 0)
   • From the calculator: binomial pdf with n = 9, p = 0.03, x = 0
   • P(X = 0) = 0.7602
   • P(X ≥ 1) = 1 − 0.7602 = 0.240

2. (b) Find the probability that the cafe returns more than 3 of the 14 cartons.

   • Y = number of cartons returned. Y ~ B(14, 0.2398…)
   • P(Y > 3) = 1 − P(Y ≤ 3)
   • From the calculator: binomial cdf with n = 14, p = 0.2398…, lower = 0, upper = 3
   • P(Y ≤ 3) = 0.5576
   • P(Y > 3) = 1 − 0.5576 = 0.442

A quiz has 40 questions, each with 5 possible answers. Omar guesses every answer at random. X is the number of questions he gets correct.

1. (a) Find the probability that Omar gets fewer than 6 correct.

   • X ~ B(40, 0.2)
   • P(X < 6) = P(X ≤ 5)
   • From the calculator: binomial cdf with n = 40, p = 0.2, lower = 0, upper = 5
   • P(X ≤ 5) = 0.161

2. (b) Find the expected number of correct answers.

   • E(X) = np = 40 × 0.2 = 8

3. (c) Find the probability that he gets more than the expected number correct.

   • P(X > 8) = P(X ≥ 9) = 1 − P(X ≤ 8)
   • From the calculator: binomial cdf with n = 40, p = 0.2, lower = 0, upper = 8
   • P(X ≤ 8) = 0.5931
   • 1 − 0.5931 = 0.407

A laboratory finds that 3% of the test tubes it puts in an oven crack. One oven load holds 50 tubes. X is the number of tubes that crack in one load.

1. (a) State two assumptions needed for X to be modelled by a binomial distribution.

   • Fixed number of trials: 50 tubes
   • Independent trials, and the probability of cracking is 0.03 for every tube

2. (b) Suggest why one of these assumptions might not hold.

   • Tubes in one load share the same oven, so a fault in the oven may make cracking more likely for several tubes
   • Then the trials are not independent

3. (c) Assume the model is valid. Find the probability that exactly 2 tubes crack.

   • From the calculator: binomial pdf with n = 50, p = 0.03, x = 2
   • P(X = 2) = 0.256

An arcade grab machine gives a prize on each try with probability 0.04, independently of other tries. Jo plays the machine again and again.

1. (a) Find the probability that at least one of the next 10 has the prize.

   • X ~ B(10, 0.04)
   • P(X ≥ 1) = 1 − P(X = 0)
   • From the calculator: binomial pdf with n = 10, p = 0.04, x = 0; P(X = 0) = 0.6648
   • P(X ≥ 1) = 1 − 0.6648 = 0.335

2. (b) Find the smallest number needed so that the probability of at least one prize is greater than 0.9.

   • X ~ B(n, 0.04). Need P(X ≥ 1) > 0.9, so P(X = 0) < 0.1
   • From the calculator: table of binomial pdf with p = 0.04, x = 0. n = 56: 0.1017. n = 57: 0.0976
   • n = 56 gives P(X ≥ 1) = 0.8983, too small. n = 57 gives 0.9024. Smallest n = 57

A basketball machine is set so that each shot scores with probability p. In 20 shots the mean number of scores is 7. X is the number of scores in 20 shots.

1. (a) Find the value of p.

   • np = 7, n = 20, so p = 7 ÷ 20 = 0.35
   • X ~ B(20, 0.35)

2. (b) Find the probability that the next 20 shots give more than 9 scores.

   • P(X > 9) = 1 − P(X ≤ 9)
   • From the calculator: binomial cdf with n = 20, p = 0.35, lower = 0, upper = 9
   • P(X ≤ 9) = 0.8782
   • P(X > 9) = 1 − 0.8782 = 0.122

X ~ B(n, p). The mean of X is 27 and the standard deviation of X is √18.9.

1. (a) Find n and p.

   • np = 27 and np(1 − p) = 18.9
   • 1 − p = 18.9 ÷ 27 = 0.7, so p = 0.3
   • n = 27 ÷ 0.3 = 90

2. (b) Find the probability that X takes the value of its mean.

   • From the calculator: binomial pdf with n = 90, p = 0.3, x = 27
   • P(X = 27) = 0.0914

A theatre has 120 seats. Each person who books a seat turns up independently with probability 0.95. The theatre accepts n bookings and X is the number of people who arrive.

1. (a) Find the mean number who arrive if 126 are sold.

   • X ~ B(126, 0.95)
   • E(X) = np = 126 × 0.95 = 119.7

2. (b) Find the largest number that can be sold so that the probability that more than 120 arrive is less than 0.05.

   • X ~ B(n, 0.95). Need P(X > 120) < 0.05, with P(X > 120) = 1 − P(X ≤ 120)
   • From the calculator: table of binomial cdf with p = 0.95, lower = 0, upper = 120, taking 1 minus each value. n = 122: 0.0142. n = 123: 0.0514
   • n = 122 is the last n below 0.05. Largest n = 122