Normal distribution

Question

Assume that the readings at freezing on a batch of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. Find P₅₁, the 51-percentile. This is the temperature reading separating the bottom 51% from the top 49%.

P₅₁ =      °C

Three decimal places.

Answer 1

0.025 degrees. You can use a table or a calculator*. What the information tells us is that The measurements are normally distributed with mean 0 and sd = 1 degree. So is our answer reasonable. We know that in a normal distribution 50% of the observations will be above and below the mean, which is 0 in this case. That's also the 50th percentile, in your notation, P₅₀.

So at just 1 more percentile point, the temperature will be just a little bit higher, so my result from a calculator or table seems quite reasonable. Using the 99.7-95-68% empirical rule, we would expect the 84th percentile to be 1 degree and we are about 1/34 of the way there, in z terms, at the 51st percentile, so I think we are good to go.

Feel free to contact me if your want to know how to get this from your particular brand and style of electronic instrument or to read your particular type of table.

*I used the function qnorm in R by the way:

> qnorm(.51,0,1)

[1] 0.02506891

Answer 2

Hi Neyda, the normal distribution specified in the problem is the standard normal distribution. P₅₁ is essentially the Z-score at which P(Z < z) = 0.51.

You can estimate this value using Z table:

http://www.z-table.com/

Best estimation from the table is P₅₁ = 0.025 deg C