Question

5.4.1 Question Help A population has a mean = 141 and a standard deviation o = 28. Find the mean and standard deviation of the sampling distribution of sample means with sample size n = 40. The mean is :-), and the standard deviation is 0;=0 (Round to three decimal places as needed.)

Answer 1

$5.4.1:~\mu_{\bar{x}}=141,~\sigma_{\bar{x}}=\frac{28}{\sqrt{40}}=4.427$

$5.4.2:~\mu_{\bar{x}}=74,~\sigma_{\bar{x}}=\frac{8}{\sqrt{64}}=1$

5.4.9: Option (6) 11.1 Mi = 17.2, Oi 0.74 225

$5.4.15:~ \mu_{\bar{x}}=22,~\sigma_{\bar{x}}=\frac{1.31}{\sqrt{68}}=0.1589.\\\\ P(\bar{X}<22.1)=\Phi((22.1-22)/0.1589)=\Phi(0.63)=0.7357;\\\\ (where,~\Phi(x)=cdf~of~N(0,1)).$

$5.4.19:~ \mu_{\bar{x}}=58,~\sigma_{\bar{x}}=\frac{6.25}{\sqrt{11}}=1.88.\\\\$

0.20 । 0.15 0.10 0.05 0.00 52 54 56 58 60 62 64 X

$5.4.19:~ \mu_{\bar{x}}=120,~\sigma_{\bar{x}}=\frac{39.7}{\sqrt{18}}=9.36.\\\\$

0.04 0.03 0.02 0.01 0.00 90 100 110 120 130 140 150 х

5. 4. 29: Required probability=0.0038 (round(pnorm(58500,61000,6000/sqrt(41)),4))

5.4.37:

(a) Required probability=0.3446 (R code: round(1-pnorm(89.24,89,0.6),4))

(b) Required probability=0.0052 (R code: round(1-pnorm(89.24,89,0.6/sqrt(41)),4))

5.4.38:

(a) Required probability=0.3569 (R code: round(1-pnorm(13.22,13,0.6),4))

(b) Required probability=0.0334 (R code: round(1-pnorm(13.22,13,0.6/sqrt(25)),4))