Question

The scores on a certain test are normally distributed with a mean score of 53 and a standard deviation of 2. What is the probability that a sample of 90 students will have a mean score of at least 53.2108? 0.8413 0.3174 0.3413 0.1587

Answer 1

$\mu=53,\;\sigma=2,\;n=90$

$P(\bar{X}\geqslant 53.2108)$

$P(\bar{X}\geqslant 53.2108)=P\left(\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt n}}\geqslant \frac{53.2108-53}{\frac{2}{\sqrt 90}}\right)$

P(X > 53.2108) = P(Z > 0.9999)

P(X > 53.2108) = 1- P(Z < 0.9999)

$P(\bar{X}\geqslant 53.2108)=1-P(Z\leqslant 1)$

$P(\bar{X}\geqslant 53.2108)=1-0.8413$

$\bold{P(\bar{X}\geqslant 53.2108)=0.1587}$

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