A metal part used in cars is manufactured with a mean diamet

A metal part used in cars is manufactured with a mean diameter of 2.565 cm and a
standard deviation of 0.008 cm. Find the probability that a random sample of n=9
sections of metal part will have a sample mean diameter greater than 2.560cm and less
than 2.570 cm.

please solve with steps

Solution

We are given that, A metal part used in cars is manufactured with a

mean diameter (mean) = 2.565 cm

standard deviation (sd) = 0.008 cm.

sample size(n) = 9

Find the probability that a random sample of n=9 sections of metal part will have a sample mean diameter greater than 2.560cm and less than 2.570 cm.

That is we have to find, P(2.560 < Xbar < 2.570)

We know that the distribution of Xbar is approximately normal with mean is 2.565 and

SD = sd / sqrt(n)

SD = 0.008 / sqrt(9) = 0.0027

Now convert Xbar into z-score.

z = (Xbar - mean) / SD

z-score for Xbar = 2.560 is,

z = (2.560 - 2.565) / 0.0027 = -1.875

z-score for Xbar = 2.570 is,

z = (2.570 - 2.565) / 0.0027 = 1.875

Now we have to find P(-1.875 < Z < 1.875).= P(Z < 1.875) - P(Z < -1.875)

We can find this probability by using EXCEL.

syntax :

=NORMSDIST(z)

where z is test statistic value.

P(Z < 1.875) = 0.9696

P(Z < -1.875) = 0.0304

P(-1.875 < Z < 1.875) = 0.9696 - 0.0304 = 0.9392