The diameter of a brand of Ping-Pong balls is approximately normally distributed

Question

The diameter of a brand of Ping-Pong balls is approximately normally distributed, with a mean of 1.32inches and a standard deviation of 0.08inch. A random sample of 4 Ping-Pong balls is selected. Complete parts (a) through (d).Click here to view page 1 of the cumulative standardized normal distribution table. Click here to view page 2 of the cumulative standardized normal distribution table.

a. What is the probability that the sample mean is less than 1.28 inches?

P(Upper X overbar less than1.28) equals.

(Round to four decimal places as needed.)

b. What is the probability that the sample is between 1.28and 1.31inches? P(1.28 less thanUpper X overbarless than1.31)

(Round to four decimal places as needed.)

c. The probability is 57% that the sample mean will be between what two values, symmetrically distributed around the population mean?

The lower bound is

(Round 4 places)

The upper bound is

(Round 4 places)

Explanation / Answer

here std error of mean =std deviation/(n)1/2 =0.08/(4)1/2 =0.04

a)  probability that the sample mean is less than 1.28 inches=P(X<1.28)=P(Z<(1.28-1.32)/0.04)=P(Z<-1) =0.1587

b) probability that the sample is between 1.28and 1.31inches=P(1.28<X<1.31)

=P((1.28-1.32)/0.04<Z<(1.31-1.32)/0.04) =P(-1<Z<-0.25)=0.4013-0.1587 =0.2426

c)for middle 57% values fall between 21.5% and 78.5%

for above z score =-/+ 0.7892

lower bound =mean +z*std deviation =1.32-0.7892*0.04 =1.2884

upper bound =mean +z*std deviation =1.32+0.7892*0.04=1.3516

please revert for any clarification required

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