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A manufacturer knows that their items have a lengths that are approximately normally distributed, with a mean of 9.8 inches, and standard deviation of 1.6 inches. If 48 items are chosen at random, what is the probability that their mean length is greater than 9.2 inches

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Answer:

0.6462

Step-by-step explanation:

Given that

Mean (m) = 9.8

Standard deviation (s) = 1.6

Probability that mean length is greater than 9.2

Z = (x - m) / s

Z = (9.2 - 9.8) / 1.6

Z = - 0.6 / 1.6

Zscore = - 0.375 = - 0.375

P(Z >- 0.375) = 1 - P(Z < - 0.375) ;

P(Z < - 0.375) = 0.3538

1 - P(Z < - 0.38) = 1 - 0.3538 = 0.6462

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