The inside diameter of a randomly selected piston ring is a random variable with

Question

The inside diameter of a randomly selected piston ring is a random variable with mean value 8 cm and standard deviation 0.03 cm. Suppose the distribution of the diameter is normal. (Round your answers to tour decimal places.) (a) Calculate P(7.99 lessthanorequalto x lessthanorequalto 8.01) when n = lb. P(7.99 lessthanorequalto x lessthanorequalto 8.01) (b) Mow likely is it that the sample mean diameter exceeds 8.01 when n = 25? P(X greaterthanorequalto 0.01) = You may need to use the appropriate table In the Appendix of Tables to answer this question.

Explanation / Answer

mean is 8 and sd is 0.03

a) for n=16, the standard error is 0.03/sqrt(16)= 0.0075

Thus P(7.99<xbar<8.01)=P((7.99-8)/0.0075<z<(8.01-8)/0.0075)=P(-1.33<z<1.33)=2*P(z<1.33)-1

from normal distribution table we get 2*0.9083-1=0.8166

b) for n=25 the standard error is 0.03/sqrt(25)=0.006

thus P(xbar>8.01)=P(z>(8.01-8)/0.006)=P(z>1.67) or 1-P(z<1.67)

from normal distribution table we get 1-0.9525 =0.0475

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