Question

Suppose xhas a distribution with 26 and 24 (round the probability to the fourth decimal place) (a) If a random sample of size n = 47 is drawn, find and P(26 SXS 28). - 35008 P(26 SXS 28) (b) if a random sample of size n = 68 is drawn, find My og and P26 SX S 28). 0 = P(26 SXS 28) = (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) is Select part (a) because of the -Select- sample sire. Therefore, the distribution about g -Select-

Answer 1

$\dpi{100} \mathbf{Solution}$

given:u= 26, o = 24

a given: n = 47

μ; =μ = 26

-r = = = 3.5008

P26 << 28) =?

$\dpi{100} \mathbf{formula:}\;Z=\frac{\bar x-\mu_{\bar x}}{\sigma_{\bar x}}$

(2) = =")。

PO<Z < 0.57)

P(Z < 0.57) - P(Z <0)

Refer Z-table to find the probability or use excel formula "=NORM.S.DIST(0.57, TRUE)" & "=NORM.S.DIST(0, TRUE)" to find the probability.

合0.7161-0.5000

-0.2161

:P 26<x<28) = 0.2161

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b) given: n = 68

μ; =μ = 26

o24 = = = 6 = 2.9104 yn

P26 << 28) =?

$\dpi{100} \mathbf{formula:}\;Z=\frac{\bar x-\mu_{\bar x}}{\sigma_{\bar x}}$

→ P (26 - 26 79104 1 - Mi, 28 – 26 2.9104

PO<Z < 0.69)

P(Z < 0.69) - P(Z <0)

Refer Z-table to find the probability or use excel formula "=NORM.S.DIST(0.69, TRUE)" & "=NORM.S.DIST(0, TRUE)" to find the probability.

合0.7540 -0.5000

一0.2540

: P26 <x<28) = 0.2540

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_{${\color{Blue} \mathbf{c)}}\;$}The standard deviation of part (b) is
smaller than
part (a) because of the
_larger
sample size. Therefore, the distribution about $\mu_{\bar x}$ is ${\color{Red} \mathbf{narrower}}$