Question

tried 0.0025 too, did not work either

A manufacturing process is designed to produce bolts with a 0.5-inch diameter. Once each day, a random sample of 36 bolts is selected and the bolt diameters recorded. If the resulting sample mean is less than 0.490 inches or greater than 0.510 inches, the process is shut down for adjustment. The standard deviation for diameter is 0.02 inches. What is the probability that the manufacturing line will be shut down unnecessarily? (Hint: Find the probability of observing an x in the shutdown range when the true process mean really is 0.5 inches. Round your answer to four decimal places.) 0.0024 You may need to use the appropriate table in Appendix A to answer this question. Need Help? Read It Talk to a Tutor

Answer 1

Solution :

Given that,

mean = $\mu$ = 0.5

standard deviation = $\sigma$ = 0.02

n = 36

$\mu$_{$\bar x$} =  $\mu$ = 0.5

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 0.02 / $\sqrt$ 36 = 0.003

P(0.490 > $\bar x$ OR $\bar x$ > 0.510)  

P($\bar x$ < 0.490) = P(($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$} < (0.490 - 0.5) / 0.003)

= P(z < -3.33)

Using z table

= 0.0004

P($\bar x$ > 0.510) = 1 - P($\bar x$ < 0.510)

= 1 - P[($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$} < (0.510 - 0.5) / 0.003]

= 1 - P(z < 3.33)

Using z table,    

= 1 - 0.9996

= 0.0004

= P($\bar x$ < 0.490 + P($\bar x$ > 0.510)

= 0.0004 + 0.0004

= 0.0008