Statistics: Finding proportions using mean and standard deviation

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I'm stuck on second part of question

so i have mean of $\mu =22.8$ and standard deviation $\sigma = 1.1$ and it is normal distribution

a) what proportion is between $22$ and $23$.

and i got $33.9\text{%}$

b) $5$ percent exceed _________

I cant figure this out... What formula would I need to use?

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2 Answers

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To figure out part b, you need to find the 95th percentile. You can use a normal table to do this, find the z value which corresponds to 95% of the data to the left of the value. In your case, this z value is 1.645. Now you can use the formula $\frac{x-\mu}{\sigma}=z=1.645$. You know that $\mu=22.8$ and $\sigma=1.1$ so we have $\frac{x-22.8}{1.1}=1.645$ thus $x=24.6095$

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For $b)$ Let $c$ be the value in question, then $P(x > c) = 0.05 \to P(x < c) = 1 - P(x > c) = 1 - 0.05 = 0.95$. And this gives $z = 1.645$. So $c - \mu = z\cdot \sigma \to c = \mu + z\cdot \sigma = 22.8 + 1.645\cdot 1.1 = 24.6$

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