A synthetic fiber used in manufacturing carpet has tensile strength that is norm

Question

A synthetic fiber used in manufacturing carpet has tensile strength that is normally distributed with mean 75 5 psi and standard deviation 3.5 psi. Find the probability that a random sample of n=6 fiber specimens will have sample mean tensile strength that exceeds 75.75 psi. The amount of time that a customer spends waiting at an airport check-in counter is a normal random variable with mean 8.2 minutes and standard deviation 1.5 minutes. Suppose that a random sample of n 49 customers is observed. Find the probability that the average time waiting in line for these customers is (a) Less than 10 minutes (b) Between 5 and 10 minutes (c) Less than 6 minutes

Explanation / Answer

mean = 75.5 , sd= 3.5

Z =(Xbar - mean)/(sd/sqrt(n))

here n = 6

hence

Z =(Xbar - 75.5)/(3.5/sqrt(6))

P(Xbar > 75.75)

= P(Z> ( (75.75 - 75.5)/(3.5/sqrt(6))

=P(Z> 0.17496355)

= 0.4306

Please ask Q2 again or it can be done similarly

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