[Math] Normal Distribution, finding the minimum mark

normal distributionstatistics

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!

Best Answer

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.

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[Math] Probability – Calculate chance total score will be over or under a score line

OK, sorry for misunderstanding your point. So, your model predicts $PV=8$ with errors $\epsilon \sim N(-0.3,13)$. Or, this is equivalent to saying that you expect the actual value to be $AV \sim N(8.3,13)$. ($PV=AV+\epsilon$)

Now, to compute the probability that $X\leq 5.5$ (or $x>5.5$) under the hypothesis that your model is correct. This is

$$\mathbb{P}(X\leq 5.5|X \sim N(8.3,13)) \; .$$

I don't know what programming language or other computer program you use, but most math or statistics interpretative languages (Matlab, R, etc...) have the necessary commands inbuilt. So in R, you would use the command

pnorm(5.5,8.3,13)

which computes the cumulative normal distribution in point $5.5$ for mean and variance as given. For the probability bigger than $5.5$, you take

1-pnorm(5.5,8.3,13)

If you want the explicit formula for the computation, I can provide that too, but unless you want to make a lot of extra work implenting the functions yourself, I don't think that is really useful.

[Math] Standard Normal Distribution $\Phi^{−1}(z)$ and $\Phi(z)$

a) is easy, we have, denoting the random variable "score" in question by $S$, that: \begin{align*} \def\P{\mathbb P}\P(S > 72) &= \P(S - 78 > -6)\\ &= \P\left(\frac{S-78}6 > -1\right)\\ &= 1 - \Phi(-1). \end{align*} b) Asks us to solve the equation $$ \P(S > s) = 0.1 $$ for $s$. We first do as in a), writing $$ \P(S > s) = \P \left(\frac{S-78}6 > \frac{s-78}6\right) = 1 - \Phi\left(\frac{s-78}6\right) $$ So, \begin{align*} 1 - \Phi\left(\frac{s-78}6\right) = 0.1 &\iff &\Phi\left(\frac{s-78}6\right) &= 0.9\\ &\iff& \frac{s-78}6 &= \Phi^{-1}(0.9)\\& \iff& s &= 78 + 6\Phi^{-1}(0.9) \end{align*}

Best Answer

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[Math] Probability – Calculate chance total score will be over or under a score line

[Math] Standard Normal Distribution $\Phi^{−1}(z)$ and $\Phi(z)$

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