A car insurance company has determined that 9% of all drivers were involved in a

Question

A car insurance company has determined that 9% of all drivers were involved in a car accident last year. Suppose that 12 drivers are randomly selected. a. Find the probability that at most three of the 12 drivers were involved in a car accident last year. b. Find the probability that at least one of the 12 drivers were involved in a ear accident last year. c. Compute the mean and standard deviation. Round the standard deviation to two decimal places. Mean = ______ Standard Deviation = ______ d. Would it be unusual to find that 4 drivers were involved in an accident last year? Justify your response.

Explanation / Answer

p = 0.09

n = 12

P(X = x) = 12Cx * 0.09x * (1 - 0.09)12-x

a) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

                  = 12C0 * 0.090 * 0.9112 + 12C1 * 0.091 * 0.9111 + 12C2 * 0.092 * 0.9110 + 12C9 * 0.099 * 0.913

                  = 0.9820

b) P(X > 1) = 1 - P(X = 0)

                   = 1 - 12C0 * 0.090 * 0.9112

                   = 1 - 0.3225

                   = 0.6775

c) Mean = n * p = 12 * 0.09 = 1.08

Standard deviation = sqrt(n * p * (1 - p)) = sqrt(12 * 0.09 * 0.91) = 0.99

d) P(X = 4) = 12C4 * 0.094 * 0.918 = 0.0153

Since the probability is less than 0.05, iy would be unusual to find that 4drivers were involved in an accident last year

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