Answer:
a. ( to four decimal places)
b. ( to four decimal places )
c. The limits will be between the interval of ( 22.33,24.67 )
Step-by-step explanation:
Given that :
mean = 23.50
standard deviation = 5.00
sample size = 50
The objective is to calculate the following:
(a) What is the likelihood the sample mean is at least $25.00?
Let X be the random variable, the probability that the sample mean is at least 25.00 is:
to two decimal places
From the normal tables :
( to four decimal places)
(b) What is the likelihood the sample mean is greater than $22.50 but less than $25.00?
to four decimal places
(c) Within what limits will 90 percent of the sample means occur?
At 90 % confidence interval, level of significance = 1 - 0.90 = 0.10
The critical value for the = 1.65
Standard Error =
Standard Error =
Standard Error = 0.7071
Therefore, at 90 percent of the sample means, the limits will be between the intervals of :
Lower limit = ( 23.5 - (1.65×0.707) )
Lower limit = ( 23.5 - 1.16655 )
Lower limit = 22.33345
Lower limit = 22.33 (to two decimal places).
Upper Limit = ( 23.5 + (1.65*0.707) )
Upper Limit = ( 23.5 + 1.16655 )
Upper Limit = 24.66655
Upper Limit = 24.67
The limits will be between the interval of ( 22.33,24.67 )
