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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?

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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, $P = 1 - (1-\hat{p})^{10} - 10 \hat{p}(1-\hat{p})^{9} = 0.8238783.$

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  • $\begingroup$ The OP question states that the average is $98.2$ while the threshold of interest is $98.6$, so this is an unfair coin. $\endgroup$ – antkam Mar 30 at 4:50
  • $\begingroup$ Oops, I didn't notice, thanks. Will fix presently... $\endgroup$ – dnqxt Mar 30 at 4:51
  • $\begingroup$ 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. :) $\endgroup$ – antkam Mar 30 at 5:01
  • $\begingroup$ Moms are usually right :) $\endgroup$ – dnqxt Mar 30 at 5:08
  • $\begingroup$ Thanks. I understand the first steps, but I thought the formula for binomial distributions was n Choose x, so how does that work? $\endgroup$ – Olivetti Apr 1 at 8:55
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Hint: You need to use both the Normal and the Binomial in different parts of the computation.

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