According to a study of 90 truckers a trucker drives on aver

According to a study of 90 truckers, a trucker drives, on average, 540 miles per day. If the standard deviation of the miles driven per day for the population of truckers is 40, find the left bound of the 99% confidence interval of the mean number of miles driven per day by all truckers. For example, if the confidence interval is 100 < < 200, the left bound is 100. Round to the nearest integer

Solution

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)=540
Standard deviation( sd )=40
Sample Size(n)=90
Confidence Interval = [ 540 ± Z a/2 ( 40/ Sqrt ( 90) ) ]
= [ 540 - 2.58 * (4.22) , 540 + 2.58 * (4.22) ]
= [ 529.12,550.88 ]

Lower Bound is 529.12 ~ 530