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

\(X \sim B(32, 0.35)\). Find

1. (a) the mean of X

   • E(X) = np = 32 × 0.35 = 11.2

2. (b) the variance of X

   • Var(X) = np(1 − p) = 32 × 0.35 × 0.65 = 7.28

3. (c) the standard deviation of X

   • SD = √7.28 = 2.70

\(Y \sim B(22, 0.35)\). Find

1. (a) \(P(Y = 8)\)

   • From the calculator: binomial pdf with n = 22, p = 0.35, x = 8
   • P(Y = 8) = 0.173

2. (b) \(P(Y > 9)\)

   • P(Y > 9) = 1 − P(Y ≤ 9)
   • From the calculator: binomial cdf with n = 22, p = 0.35, lower = 0, upper = 9
   • P(Y ≤ 9) = 0.7916
   • P(Y > 9) = 1 − 0.7916 = 0.208

3. (c) \(P(5 \le Y < 9)\)

   • P(5 ≤ Y < 9) = P(5 ≤ Y ≤ 8)
   • From the calculator: binomial cdf with n = 22, p = 0.35, lower = 5, upper = 8
   • P(5 ≤ Y ≤ 8) = 0.575

Each of 6 seedlings independently survives a frost with probability 0.3. X is the number of seedlings that survive.

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

   • X ~ B(6, 0.3)
   • From the calculator: binomial pdf with n = 6, p = 0.3, x left blank gives P(X = 0) to P(X = 6)
   • x = 0 to 6: 0.118, 0.303, 0.324, 0.185, 0.0595, 0.0102, 0.000729
   • The total is 1, apart from rounding

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

   • The largest probability is 0.324, at x = 2

A pack holds 8 pens. Each pen is faulty with probability 0.04, independently. A school buys 12 packs and returns any pack that has at least one faulty pen.

1. (a) Find the probability that a pack has at least one faulty pen.

   • X = number of faulty pens in a pack of 8 pens. X ~ B(8, 0.04)
   • P(X ≥ 1) = 1 − P(X = 0)
   • From the calculator: binomial pdf with n = 8, p = 0.04, x = 0
   • P(X = 0) = 0.7214
   • P(X ≥ 1) = 1 − 0.7214 = 0.279

2. (b) Find the probability that the school returns more than 2 of the 12 packs.

   • Y = number of packs returned. Y ~ B(12, 0.2786…)
   • P(Y > 2) = 1 − P(Y ≤ 2)
   • From the calculator: binomial cdf with n = 12, p = 0.2786…, lower = 0, upper = 2
   • P(Y ≤ 2) = 0.3075
   • P(Y > 2) = 1 − 0.3075 = 0.693

A quiz has 28 questions, each with 4 possible answers. Priya guesses every answer at random. X is the number of questions she gets correct.

1. (a) Find the probability that Priya gets fewer than 5 correct.

   • X ~ B(28, 0.25)
   • P(X < 5) = P(X ≤ 4)
   • From the calculator: binomial cdf with n = 28, p = 0.25, lower = 0, upper = 4
   • P(X ≤ 4) = 0.135

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

   • E(X) = np = 28 × 0.25 = 7

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

   • P(X > 7) = P(X ≥ 8) = 1 − P(X ≤ 7)
   • From the calculator: binomial cdf with n = 28, p = 0.25, lower = 0, upper = 7
   • P(X ≤ 7) = 0.5997
   • 1 − 0.5997 = 0.400

A courier says that 4% of the parcels it delivers arrive damaged. A van carries 60 parcels. X is the number of damaged parcels.

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

   • Fixed number of trials: 60 parcels
   • Independent trials, and the probability of damage is 0.04 for every parcel

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

   • Parcels in one van are loaded and driven together, so damage to one may make damage to another more likely
   • Then the trials are not independent

3. (c) Assume the model is valid. Find the probability that exactly 3 parcels are damaged.

   • From the calculator: binomial pdf with n = 60, p = 0.04, x = 3
   • P(X = 3) = 0.214

A packet of cereal contains a free toy with probability 0.06, independently of other packets. Sam buys packets one after another.

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

   • X ~ B(10, 0.06)
   • P(X ≥ 1) = 1 − P(X = 0)
   • From the calculator: binomial pdf with n = 10, p = 0.06, x = 0; P(X = 0) = 0.5386
   • P(X ≥ 1) = 1 − 0.5386 = 0.461

2. (b) Find the smallest number needed so that the probability of at least one prize is greater than 0.95.

   • X ~ B(n, 0.06). Need P(X ≥ 1) > 0.95, so P(X = 0) < 0.05
   • From the calculator: table of binomial pdf with p = 0.06, x = 0. n = 48: 0.0513. n = 49: 0.0482
   • n = 48 gives P(X ≥ 1) = 0.9487, too small. n = 49 gives 0.9518. Smallest n = 49

A biased spinner lands on red with probability p. In 15 spins the mean number of reds is 9. X is the number of reds in 15 spins.

1. (a) Find the value of p.

   • np = 9, n = 15, so p = 9 ÷ 15 = 0.6
   • X ~ B(15, 0.6)

2. (b) Find the probability that the next 15 spins give fewer than 6 reds.

   • P(X < 6) = P(X ≤ 5)
   • From the calculator: binomial cdf with n = 15, p = 0.6, lower = 0, upper = 5
   • P(X ≤ 5) = 0.0338

X ~ B(n, p). The mean of X is 20 and the standard deviation of X is √12.

1. (a) Find n and p.

   • np = 20 and np(1 − p) = 12
   • 1 − p = 12 ÷ 20 = 0.6, so p = 0.4
   • n = 20 ÷ 0.4 = 50

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

   • From the calculator: binomial pdf with n = 50, p = 0.4, x = 20
   • P(X = 20) = 0.115

A ferry has room for 50 cars. Each driver who books turns up independently with probability 0.9. The company accepts n bookings and X is the number of cars that arrive.

1. (a) Find the mean number who arrive if 55 are sold.

   • X ~ B(55, 0.9)
   • E(X) = np = 55 × 0.9 = 49.5

2. (b) Find the largest number that can be sold so that the probability that more than 50 arrive is less than 0.05.

   • X ~ B(n, 0.9). Need P(X > 50) < 0.05, with P(X > 50) = 1 − P(X ≤ 50)
   • From the calculator: table of binomial cdf with p = 0.9, lower = 0, upper = 50, taking 1 minus each value. n = 52: 0.0283. n = 53: 0.0898
   • n = 52 is the last n below 0.05. Largest n = 52