Question

with ?-23 and ? " 15. (a) If a random sample of size n-32 is drawn, find AG, ?i and P 23 S x s 25). (Round to two decimal places and the probability to four decimal places.) P(23 s x s 25) (b) If a random sample of size n-60 is drawn, rind Min ?i and ,r23 s x s 25). (Round ?? to two decimal places and the probability to four decimal places.) P(23 s x s 25) (c) Why should you expect the probability of part (b) to be higher than that of part (a)? (Hint: Consider the standard deviations in parts (a) and (b).) The standard deviation of part (b) islct part (a) because of the Select sample size. Therefore, the distribution about j is -Select-

Answer 1

1) Mean =23 and standard deviation=15

a) n=32

$\sigma _X=\sigma /\sqrt{n}=15/\sqrt{32}=2.6517$

$P(23\leq X\leq 25)=P(\frac{23-23}{2.6517}\leq Z\leq \frac{25-23}{2.6517})$

$P(0\leq Z\leq 0.75425)=0.2746$

b) n=60

$\mu _X=23$

$\sigma _X=\sigma /\sqrt{n}=15/\sqrt{60}=1.9365$

$P(23\leq X\leq 25)=P(\frac{23-23}{1.9365}\leq Z\leq \frac{25-23}{1.9365})$

$P(0\leq Z\leq 1.032796)=0.3492$

c)The standard deviation of part(b) is lower because of the higher sample size.