Question

distributed, with a mean of 1.31 inches and a standard deviation of 0.04 inch A random sample The dameter of a brand of ping-pong ball is approximately normally balls is seledted Complete parts (a) through (d) of 16 ping-pong is a. What is the sampling distribution of the mean? O A. Because the population dameter of Ping Pong bals s approxdmahely nomaly distrbuted, the sampling distbution of samples of 16 will be the uniform distribution OB. Because the population dan eser of p ngp ong bans is appro inatnay normally distributed,the sampling dishbutin of samples of 16 wil also be approximabely nomal O C. Because the population dameter of Ping Pong balls ih apprexsimately normally dstrbuted, he sampling distribution of samples of 16 can not be found O D. Because the population dameter of Ping-Pong balls is approximatoly nomally dstribuled, the samplng distriution of samples of 16 will not be approximaterly normal b. What is the probability that the sample mean is less than 1 28 inches? Round to four decienal places as needed) с.what is the petabay eatthe sangle is between 12S and 134 inches? Round to Youar decimal places as needed) The lower bound is.๒dws Round to er decimal places an needed) heper bound nce Round to twe deciemall ylaces as needed 5 7 as lock

Answer 1

(a)

Correct option is B.

(b)

The z-score for $\bar{x}=1.28$ is

$z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\frac{1.28-1.31}{0.04/\sqrt{16}}=-3$

The required probability is

$P(\bar{x}\leq 1.28)=P(z\leq -3)=0.0013$

(c)

The z-score for $\bar{x}=1.34$ is

$z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\frac{1.34-1.31}{0.04/\sqrt{16}}=3$

The required probability is

$P(1.28<\bar{x}<1.34)=P(-3<z<3)=0.9973$

(d)

We need z-scores that have 0.58 area between them. The z-scores -0.806 and 0.806 have 0.58 area between them.

The lower bound is

$-0.806=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\frac{\bar{x}-1.31}{0.04/\sqrt{16}}$

$\bar{x}\approx 1.30$

The upper bound is

$0.806=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\frac{\bar{x}-1.31}{0.04/\sqrt{16}}$

- 1.32