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Suppose the score of each student in an exam follows a Normal distribution with mean $\mu=75$ and standard deviation $\sigma=0.8$. If we take a random sample of 20 of these students, then what is the probability that more than 2 students will score less than 74.4

I know that for a sample size $n$, the mean of the random sample remains the same ($\mu$) as the population but the standard deviation becomes $\frac{\sigma}{\sqrt{n}}$.

The thing which is confusing me here is this part: probability that more than 2 students will score less than 74.4

If they would have asked about the probability of the mean score of the 20 randomly selected samples being less than 74.4, then the solution was simple. This is what I would have done

$$ z = \frac{74.4-75}{\frac{0.8}{\sqrt{20}}}=-3.354 \\ P(z<-3.354) = 0.0004 $$

But how to deal with the situation which is asked in the original question. Any help would be appreciated.

Thanks in advance :)

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  • $\begingroup$ In R: code 1 - pbinom(1, 20, pnorm(74.4, 75, .8)) returns $0.9598072.$ $\endgroup$
    – BruceET
    Apr 6, 2022 at 16:10

1 Answer 1

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Hint:

  1. find the probability of one individual getting a score of 74.4 or less,
  2. use the binomial distribution to incorporate sample size.
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