Consider the following binomial random variables. (a) The number of tails seen i

Question

Consider the following binomial random variables. (a) The number of tails seen in 52 tosses of a quarter. (i) Find the mean. (Give your answer correct to one decimal place.) (ii) Find the standard deviation. (Give your answer correct to two decimal places.) (b) The number of left-handed students in a classroom of 45 students (assume that 12% of the population is left-handed). (i) Find the mean. (Give your answer correct to one decimal place.) (ii) Find the standard deviation. (Give your answer correct to two decimal places.) (c) The number of cars found to have unsafe tires among the 421 cars stopped at a roadblock for inspection (assume that 16% of all cars have one or more unsafe tires). (i) Find the mean. (Give your answer correct to one decimal place.) (ii) Find the standard deviation. (Give your answer correct to two decimal places.) (d) The number of melon seeds that germinate when a package of 42 seeds is planted (the package states that the probability of germination is 0.87. (i) Find the mean. (Give your answer correct to one decimal place.) (ii) Find the standard deviation. (Give your answer correct to two decimal places.) Need Help?

Explanation / Answer

a) n = 52

P(tail), p = 0.5

Mean = n * p = 52 * 0.5 = 26

Standard deviation = sqrt(n * p * (1 - p)) = sqrt(52 * 0.5 * 0.5) = 3.61

b) n = 45

p = 0.12

Mean = n * p = 45 * 0.12 = 5.4

Standard deviation = sqrt(n * p * (1 - p)) = sqrt(45 * 0.12 * 0.88) = 2.18

c) n = 421

p = 0.16

Mean = n * p = 421 * 0.16 = 67.36

Standard deviation = sqrt(n * p * (1 - p)) = sqrt(421 * 0.16 * 0.84) = 7.52

d) n = 42

p = 0.87

Mean = n * p = 42 * 0.87 = 36.54

Standard deviation = sqrt(n * p * (1 - p)) = sqrt(42 * 0.87 * 0.13) = 2.18

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