# 130 healthy people are selected at random, what is the probability that the average temperature for these people is 98.25 degrees or lower?Assume that temperatures of humans isapproximately...

130 healthy people are selected at random, what is the probability that the average temperature for these people is 98.25 degrees or lower?

Assume that temperatures of humans is

approximately normal with a mean of 98.6 degrees and

a st dev of 0.8 degrees.

### 1 Answer | Add Yours

Given `mu=98.6,sigma=0.8` (The population mean temperature is 98.6 and the population standard deviation is 0.8) and assuming the populationis approximately normal:

Find the probability that in a random sample of 130 people, the average temperature is 98.25 or lower.

We want to find `P(bar(x)<=98.25)` . We need to convert this to a normalized score. We use `z=(bar(x)-mu)/(sigma/sqrt(n))` : The population is normal and we know the population standard deviation so we use the `z` score, and we need to accoount for the fact that a sample has less variability than the population so we use the standard error `sigma/sqrt(n)` .

Then `z=(98.25-98.6)/(.8/sqrt(130))~~-4.988`

Thus `P(bar(x)<=98.25)=P(z<-4.99)` .

If you consult the standard normal table for `z<-4.99` you will find that for `z<-3.5` you use .0001. (My calculator returns .0000003)

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**The probability that a sample of 130 people has a mean temperature of 98.25 or less is .0001**

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