A company makes a certain automotive engine part called a \'kumquat\'. The distr

Question

A company makes a certain automotive engine part called a 'kumquat'. The distribution of the time for engines using this part to accelerate from 0 to 60 mph is found to be Normally distributed with a mean of 7.3 seconds and a standard deviation of 2.1 seconds. Suppose you take a random sample of 50 engines using this part and measure their acceleration time (i.e. to go from 0 to 60). a) What would be the mean and standard deviation of the distribution of sample means? b) What is the probability of coming up with a sample that shows an acceleration time of less than 7 seconds?

Explanation / Answer

Solution:- Given data mean = 7.3 seconds ,  standard deviation = 2.1 seconds , sample n = 50

a)  the mean and standard deviation of the distribution of sample means :

x = 7.3 seconds

x = /sqrt(n) = 2.1/sqrt(50) = 0.2970

b) P(X < 7) = P(Z < (X - )//sqrt(n) )

= P(Z < (7 - 7.3)/2.1/sqrt(50) )

= P(Z < -0.0202)

= 0.4920

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