Probability of distributions

Question

A particular question says the average body temperature is 98.2F with standard deviation of 0.7 with normal distribution, allowing us to find probability with z-tables.

However, an additional question asks that if 10 people were selected at random (independent of each other) and X represents the number of people who have body temperatures exceeding 98.6F. What is the probability that at least 2 people have temperatures that exceed 98.6F?

Would this distribution still be Normal, as the parameters of mean and SD still apply, or is it binomial? As there is a fixed amount of trials and success/failure as limits. But then the parameters for n and p, what would be p?

Answer 1

Hint: You need to use both the Normal and the Binomial in different parts of the computation.

Answer 2

Given the normal distribution, $N(98.2, 0.7^2)$, the probability of someone having temp higher than 98.6 is $\hat{p}=1-\Phi(\frac{98.6-98.2}{0.7})= 0.2838546$.

Sample size is $n=10$ and the probability of at least two people having higher temp is: $P = 1 - P(X=0) - P(X=1)$, where $X$ is distributed as Binomial$(n,\hat{p})$. Thus, $P = 1 - (1-\hat{p})^{10} - 10 \hat{p}(1-\hat{p})^{9} = 0.8238783.$