Using the R script solve the following: An expert on process control states that he is 95% confident that the new production process will save between $26 and $38 per unit with savings values around $32 more likely. If you were to model this expert's opinion using a normal distribution (by applying empirical rule), what standard deviation would you use for your normal distribution? (round your answer to 1 decimal place.

Can someone suggest what is the correct method of solving this problem? Please provide R script

I used 2 methods:

METHOD 1: Found Z score using qnorm(0.95)

This comes to 1.644854

The population mean then is 32

the mean + 1.644854 standard deviations is 38 (95% of customers save no more than this)

38 - 32 = 6 (this is equal to 1.644854 StDev)

6 = 1.644854 * stdev

6/1.644854 = StDev

StDev = 3.64774 (expected answer is to be rounded to one decimal)

Then I verified that I get the upper bound by using: qnorm(.95,mean=32,sd=3.64774) = 38

However, this is not the right answer

METHOD 2:

According to Empirical rule,95% of the data falls **within 2 standard deviations of the mean**

upper limit=38=32+2*sd
6=2*sd

sd=3

When I plug this sd in qnorm(.95,mean=32,sd=3.0), I get a value of 36.93 and not 38