The weekly earning of bus drivars are normally distributed w

The weekly earning of bus drivars are normally distributed with a ,eam of mu=$395. If only 1.1% of the bus drivers have a weekly income of more than $429.35, what is the value of the variance of the weekly earning of the bus drivers? A random sample of n=100 credit sales in a department store showed an average sale of . From past data, it is known that the standard deviation of the population is sigma=$40.00. Find the 95% confidence interval of the population mean. Interpret this interval

Solution

a.
Margin of Error = Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
Mean(x)=120
Standard deviation( sd )=40
Sample Size(n)=100
Margin of Error = Z a/2 * 40/ Sqrt ( 100)
= 1.96 * (4)
= 7.84

b.Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=120
Standard deviation( sd )=40
Sample Size(n)=100
Confidence Interval = [ 120 ± Z a/2 ( 40/ Sqrt ( 100) ) ]
= [ 120 - 1.96 * (4) , 120 + 1.96 * (4) ]
= [ 112.16,127.84 ]