Probability of distributions

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?

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, $$P = 1 - (1-\hat{p})^{10} - 10 \hat{p}(1-\hat{p})^{9} = 0.8238783.$$
• The OP question states that the average is $98.2$ while the threshold of interest is $98.6$, so this is an unfair coin. – antkam Mar 30 at 4:50
• I dont blame you. My mom has always told me the average is $98.6$, i.e. my mom disagrees with the premise of this question. :) – antkam Mar 30 at 5:01