The diameter of a brand of PingPong balls is approximately n

The diameter of a brand of Ping-Pong balls is approximately normally distributed, with a mean of 1.31 inches and a standard deviation of 0.08 inch. A random sample of 4 Ping-Pong balls is selected. Complete parts (a) through (d). what is the sampling distribution of the mean? Because the population diameter of Ping-Pong balls is approximately normally distributed, the sampling distribution of samples of 4 will not be approximately normal. Because the population diameter of Ping-Pong balls is approximately normally distributed, the sampling distribution of samples of 4 will also be approximately normal Because the population diameter of Ping-Pong balls is approximately normally distributed, the sampling distribution of samples of 4 can not be found.

Solution

Given X~Normal(mean=1.31, s=0.08)
n=4
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(b) P(xbar<1.27)

= P((xbar-mean)/(s/n) < (1.27-1.31)/(0.08/2))

=P(Z<-1)

= 0.1587 (check standard normal table)

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(c) P(1.27<X<1.32)

=P((1.27-1.31)/(0.08/2) <Z< (1.32-1.31)/(0.08/2))

=P(-1<Z<0.25)

=0.44 (check standard normal table)

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(d)

P(-k<Z<k) = 0.6

2*P(Z<k) - 1 = 0.6

P(Z<k) = 0.8

from table k = 0.7881

k = (x-u)/s

x = 1.37

-k = (x-u)/s

x = 1.25

lower bound = 1.25

upper bound = 1.37