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

(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

Explanation:

Given -

Mean
`(\\u )` = 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} - \\u}{(\sigma)/(√(n))} > (59 - 55 )/((14)/(√(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} - \\u}{(\sigma)/(√(n))} < (54 - 55 )/((14)/(√(25))))`

=
`P(Z< -.357 )`

= .3632

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

`P(54< \overline{X}< 59)` =
`P((54 - 55 )/((14)/(√(25)))< \frac{\overline{X} - \\u}{(\sigma)/(√(n))} < (59 - 55 )/((14)/(√(25))))`

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

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

= .9222 - .3632

= 0.559