22 credit card holders are selected at random For each their

22 credit card holders are selected at random. For each, their current credit card balance is recorded. The average for these 22 people is = $600. Assume that the current balance of all credit card holders follows a normal distribution with unknown mean , and that a 85.3% confidence interval for is found to be $600 ± 30.2. Find the standard deviation of the population?

Solution

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)=600
Standard deviation( sd )= Unknown
Sample Size(n)=22
Confidence Interval = [ 600 ± Z a/2 ( Unknown / Sqrt ( 22) ) ]
[ 600 ± 30.2 ] = [ [ 600 ± 1.45 * ( Unknown / Sqrt ( 22) ) ]

By equality the second part

30.2 = 1.45 * ( Unknown / Sqrt ( 22) )

( Unknown / Sqrt ( 22) ) = 30.2/1.45 = 20.82

Unknown = 20.82 *Sqrt ( 22) = 97.6544

Standard deviation is 97.6544