# Calculate Standard Deviation with just mean and cumulative probability

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+2sd
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

One suggestion: what is the value of `qnorm(0.975)` and why should you care about it?

Actually, qnorm(0.975) is not needed.
I used this code:

f <- function(lwr, upr){
c("mean"= (upr+lwr)/2,
"stddev" = (upr-lwr)/4,
"sdRound" =round((upr-lwr)/4,1)
)
}
f(2,3)

With this, I get the answers as:
mean stddev sdRound
2.50 0.25 0.20

I can't use the value rounded in R. The correct answer is 0.3 since 0.25 when rounded is 0.3.
When I plug this sd=0.3 in the below, I get the correct upperlimit (and also lower limit)

upperlimit = round(qnorm(0.95, mean=2.5, sd=0.3),0)
upperlimit =3
lowerlimit = round(qnorm(0.05, mean=2.5, sd=0.3))
lowerlimit =2

It works for smaller values of upper and lower limits
f(2,3)
f(6,9)
but for pairs like f(6,14). f(26,38), I get the correct lowerlimit and upperlimit values only if I add 0.5 to the sd derived using the function.

Not sure if this is the correct way of solving this.

Usually these problems involve samples of n items, but let's ignore that.
A 95% confidence interval equals ( X - 1.96SD, X + 1.96SD ).
Here X = 32, upper limit = 38.
38 = 32 + 1.96*SD.
SD = (38 - 32)/1.96 = 3.06

Note that the 1.96 is the result of qnorm(.975, 0, 1) which is the cumulative distribution at .975, not at .95

1 Like

What fcas80 is saying. I was trying to bring your attention on the fact that, if the lower limit is at `0.05` and the upper at `0.95`, then you have 5% below the lower and 5% above the upper, so you're computing the 90% confidence interval.

Ahhhh, I see now. Thank you, Alexis and fcas80!

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