The breaking strength of a rivet has a mean value of 10,000 psi and a standard deviation of 500 psi.
a. What is the probability that the sample mean breaking strength for a random sample of 40 rivets is between 9950 and 10,250?
b. If the sample size had been 15 rather than 40, could the probability requested in part (a) be calculated from the given information?

Question

Quizplus · Accepted Answer

The Answer is

\mu = 10,000 \mathrm { psi } , \sigma = 500 \mathrm { psi }

a. n = 40

P ( 9,950 \leq \bar { X } \leq 10,250 ) \approx P \left( \frac { 9,950 - 10,000 } { 500 / \sqrt { 40 } } \leq Z \leq \frac { 10,250 - 10,000 } { 500 / \sqrt { 40 } } \right)

= P ( - .63 \leq Z \leq 3.16 )

= \Phi ( 3.16 ) - \Phi ( - .63 )

=.9992 - .2643 =.7349
b. According to the Rule of Thumb given in your text, n should be greater than 30 in order to
apply the C.L.T., thus using the same procedure for n = 15 as was used for n = 40 would not appropriate.

The Breaking Strength of a Rivet Has a Mean Value