Question

A normal population has a mean of 55 and a standard deviation of 14. You select a random sample of 25. Compute the probability that the sample mean is: (Round your z values to 2 decimal places and final answers to 4 decimal places): Greater than 59. Less than 54. Between 54 and 59.

Answer 1

Answer:

(1) The probability that the sample mean  is  Greater than 59 = 0.778

(2) The probability that the sample mean  is  less than 59 = .3632

(3) The probability that the sample mean  is  is Between 54 and 59 = 0.559

Step-by-step explanation:

Given -

Mean $(\nu )$ = 55

Standard deviation $(\sigma )$ = 14

Sample size ( n ) = 25

the probability that the sample mean $(\overline{X})$  is  Greater than 59 =

$P(\overline{X} > 59 )$   =  $P(\frac{\overline{X} - \nu}{\frac{\sigma}{\sqrt{n}}} > \frac{59 - 55 }{\frac{14}{\sqrt{25}}})$

                 =  $P(Z > 1.428 )$

                  = 1 - $P(Z < 1.428 )$

                  =  1 - 0.9222 = .0778

the probability that the sample mean $(\overline{X})$ is  Less than 54 =

$P(\overline{X}< 54 )$   = $P(\frac{\overline{X} - \nu}{\frac{\sigma}{\sqrt{n}}} < \frac{54 - 55 }{\frac{14}{\sqrt{25}}})$

                    = $P(Z< -.357 )$

                    =  .3632

the probability that the sample mean is Between 54 and 59 =

$P(54< \overline{X}< 59)$ = $P(\frac{54 - 55 }{\frac{14}{\sqrt{25}}}< \frac{\overline{X} - \nu}{\frac{\sigma}{\sqrt{n}}} < \frac{59 - 55 }{\frac{14}{\sqrt{25}}})$

                         = $P(-.357< Z< 1.428)$

                          = $P(Z< 1.428 ) - P(< -.357)$

                          = .9222 - .3632

                           = 0.559