AI SL · 13.2
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Mean, variance and binomial practice
13.2.6
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n trials, each a success with probability p. X is the number of successes. Choose which values of X to add up, or press a bar.
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Move a slider to change the distribution.
A library finds that each book borrowed is returned late with probability 0.18, independently. Of 45 books borrowed, X are returned late. Find the mean and the standard deviation of X.
Have a go first. Then open the working one line at a time.
X ~ B(45, 0.18)
n = 45, p = 0.18.
E(X) = np = 45 × 0.18 = 8.1
The mean need not be a whole number.
Var(X) = np(1 − p) = 45 × 0.18 × 0.82 = 6.642
Multiply by 1 − p as well.
SD = √6.642 = 2.58
The standard deviation is the square root of the variance.
Each of 5 students independently arrives late to a lesson with probability 0.4. X is the number of students who arrive late. Write down the probability distribution of X as a table, and find P(X = 3).
Have a go first. Then open the working one line at a time.
X ~ B(5, 0.4)
X takes the values 0 to 5.
From the calculator: binomial pdf with n = 5, p = 0.4, x left blank gives P(X = 0) to P(X = 5)
One pdf with x left blank gives all six probabilities.
x = 0 to 5: 0.0778, 0.259, 0.346, 0.230, 0.0768, 0.0102
The bar chart above shows the same values.
The total is 1. P(X = 3) = 0.230
A distribution always adds up to 1.
A pack holds 12 batteries. Each battery is a dud with probability 0.02, independently. (a) Find the probability that a pack has at least one dud. (b) A shop buys 15 packs. A pack counts as failed if it has at least one dud. Find the probability that at least 2 of the 15 packs fail. The box asks for part (a).
Have a go first. Then open the working one line at a time.
X = number of duds in a pack. X ~ B(12, 0.02)
(a) 12 trials, p = 0.02.
P(X ≥ 1) = 1 − P(X = 0)
At least one is everything except none.
From the calculator: binomial pdf with n = 12, p = 0.02, x = 0; P(X = 0) = 0.7847 so P(X ≥ 1) = 1 − 0.7847 = 0.215
Keep the unrounded value for part (b).
Y = number of failed packs. Y ~ B(15, 0.2153…)
(b) Each pack fails with probability 0.2153…, independently. This is a second binomial built on the first.
P(Y ≥ 2) = 1 − P(Y ≤ 1); From the calculator: binomial cdf with n = 15, p = 0.2153…, lower = 0, upper = 1; P(Y ≤ 1) = 0.1348, so P(Y ≥ 2) = 1 − 0.1348 = 0.865
Use the stored, unrounded value for p.
A scratch card wins a prize with probability 0.08, independently of other cards. Mia buys n cards. Find the smallest value of n for which the probability that she wins at least one prize is greater than 0.9.
Have a go first. Then open the working one line at a time.
X = number of winning cards. X ~ B(n, 0.08) Need P(X ≥ 1) > 0.9
n is the unknown, so there is no single calculation to type.
P(X ≥ 1) = 1 − P(X = 0) > 0.9, so P(X = 0) < 0.1
None is easier to find than at least one.
From the calculator: table of binomial pdf with p = 0.08, x = 0, for increasing n. n = 27: P(X = 0) = 0.1053. n = 28: P(X = 0) = 0.0968
Look for the first n where the value drops below 0.1.
n = 27: P(X ≥ 1) = 0.8947 (too small). n = 28: P(X ≥ 1) = 0.9032. Smallest n = 28
The chance only rises as n grows, so the first n that works is the smallest.
A coach has 80 seats. Each person who buys a ticket turns up independently with probability 0.9. The company sells n tickets. Find the largest n for which the probability that more than 80 people turn up is less than 0.1.
Have a go first. Then open the working one line at a time.
X = number who turn up. X ~ B(n, 0.9) Need P(X > 80) < 0.1
The n in the distribution is the unknown.
P(X > 80) = 1 − P(X ≤ 80)
More than 80 is everything except 80 or fewer.
From the calculator: table of binomial cdf with p = 0.9, lower = 0, upper = 80, for increasing n. n = 84: P(X > 80) = 0.0264. n = 85: 0.0643. n = 86: 0.1288
The risk rises with n. Look for the last n below 0.1.
n = 85
85 tickets are the most the company can sell and keep the risk below 10%.
X ~ B(24, 0.35). Find the probability that X is greater than its mean.
Have a go first. Then open the working one line at a time.
E(X) = np = 24 × 0.35 = 8.4
Find the mean first.
P(X > 8.4) = P(X ≥ 9) = 1 − P(X ≤ 8)
X is a whole number, so X greater than 8.4 means 9 or more.
From the calculator: binomial cdf with n = 24, p = 0.35, lower = 0, upper = 8
Lower 0, upper 8.
P(X ≤ 8) = 0.5257, so P(X ≥ 9) = 1 − 0.5257 = 0.474
A seed company tests batches of 40 seeds. On average 28 seeds in a batch germinate. Assume X, the number that germinate, is binomial. Find the probability that fewer than 25 seeds germinate in the next batch.
Have a go first. Then open the working one line at a time.
np = 28, n = 40, so p = 28 ÷ 40 = 0.7
The average is the mean, np.
X ~ B(40, 0.7)
Now n and p are both known.
P(X < 25) = P(X ≤ 24); From the calculator: binomial cdf with n = 40, p = 0.7, lower = 0, upper = 24
Lower 0, upper 24.
P(X ≤ 24) = 0.115
X ~ B(n, p). The mean of X is 14 and the standard deviation of X is √9.1. Find the probability that X takes the value of its mean.
Have a go first. Then open the working one line at a time.
np = 14 and np(1 − p) = 9.1
The variance is the standard deviation squared: 9.1.
1 − p = 9.1 ÷ 14 = 0.65, so p = 0.35
Dividing the second by the first removes n.
n = 14 ÷ 0.35 = 40
X ~ B(40, 0.35).
From the calculator: binomial pdf with n = 40, p = 0.35, x = 14; P(X = 14) = 0.131
A quiz has 40 questions, each with 4 possible answers. Sam guesses every answer at random. Find the probability that he gets fewer than 8 correct, and state the assumptions you made.
Have a go first. Then open the working one line at a time.
X = number correct. p = 1/4 = 0.25, X ~ B(40, 0.25)
Each guess has 4 equally likely answers, one right.
Each guess is made at random; guesses are independent; every question has exactly one correct answer out of 4.
These are the binomial conditions in the context of the quiz.
P(X < 8) = P(X ≤ 7); From the calculator: binomial cdf with n = 40, p = 0.25, lower = 0, upper = 7
Lower 0, upper 7.
P(X ≤ 7) = 0.182
If Sam knew some answers, p would be higher: the model would not fit.