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The exam scores on a certain Society of Actuaries (SOA) professonal examination are Normally distributed with a mean score of $μ=65\%$ and a standard deviation of σ=6

$35\%$ of all persons writing this SOA Examination will not pass. What is the minimum mark needed to pass this exam? Enter your answer to two decimal places, and enter as a percentage.

I tried to enter P(X less than Xo) = 0.35 and came out with $62.69$ ;(which is wrong)

I need some help on what to do

Thanks!

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You have percentage scores $X \sim \mathsf{Norm}(\mu = 65,\, \sigma=6)$ and you seek $c$ such that $P(X \le c) = .35.$ This can be solved by standardizing and using printed tables of the standard normal distribution:

$$.35 = P(X < 35) = p\left(\frac{X - \mu}{\sigma} < \frac{c - 65}{6}\right) = P\left(Z < \frac{c - 65}{6}\right),$$

where $Z$ is standard nomal. From tables you should be able to find that $(c-65)/6 \approx -0.38,$ and then it is a simple matter to solve for $c,$ and round up to the nearest integer (percentage) and thus get the lowest passing score.

If you have access to statistical software or a statistical calculator, you may be able to solve the question directly, without standardizing. In R statistical software qnorm is the quantile function (inverse CDF) of a normal distribution. Thus the lowest passing score is 63.

qnorm(.36, 65, 6)
## 62.84925

I find that drawing a sketch is insurance against getting tangled up reading printed normal tables and finding an "answer" that is obviously wrong. By hand, you can't make a sketch as accurate as the plot below, but you can make a sketch that is better than nothing.

enter image description here

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  • $\begingroup$ Thanks for answer @BruceET, but for some reason when I enter that answer into my assignment, it came out incorrect -- I tried to re-enter this into the software and the answer came out to be 62.69. $\endgroup$ – Gill Dave Nov 13 '17 at 1:03
  • $\begingroup$ I was assuming you were to get the nearest integer score, as at the end of your first paragraph. With normal tables I doubt you can get accuracy beyond the nearest integer pct. // Earlier you said you got 62.69 and that is wrong. // Seems one or both of us is confused. // Anyhow, I'm pretty sure the basic method I showed you is right. Maybe you can find a typo in my work or yours. $\endgroup$ – BruceET Nov 13 '17 at 1:25

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