Question

industry standards
suggest that 10% of new vehicles require warranty service within
the first year. Jones Nissan, sold 9 Nissans yesterday.
**(Round the Mean answer to 2 decimal places and the other
answers to 4 decimal places.)**

**a.**
What is the probability that none of these vehicles requires
warranty service?

Probability

**b.**
What is the probability that exactly one of these vehicles requires
warranty service?

Probability

**c.**
Determine the probability that exactly two of these vehicles
require warranty service.

Probability

**d.**
What is the probability that less than three of these vehicles
require warranty service?

Probability

**e.**
Compute the mean and standard deviation of this probability
distribution.

Mean µ = | |

Standard deviation σ = |

Answer #1

This is a binomial distribution.

n = 9

p = 0.1

P(X = x) = 9Cx * 0.1^{x} * (1 - 0.1)^{9-x}

a) P(X = 0) = 9C0 * 0.1^{0} * 0.9^{9} =
0.3874

b) P(X = 1) = 9C1 * 0.1^{1} * 0.9^{8} =
0.3874

c) P(X = 2) = 9C2 * 0.1^{2} * 0.9^{7} =
0.1722

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

= 0.3874 + 0.3874 + 0.1722

= 0.9470

e) Mean = n * p = 9 * 0.1 = 0.9

Standard deviation = sqrt(n * p * (1 - p)) = sqrt(9 * 0.1 * 0.9) = 0.9

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