Normal Distribution Statistics

Question

I am trying to understand normal distribution, and I am trying to get

Verbal SAT scores following the normal (430, 100) distribution. What is the middle range of scores encompassing 50% of the population?

My attempt :

Since X~normal(430,100) denote SAT score, we want y such that $P(\frac{430-y-430}{10} \le Z \le \frac{430+y-430}{10}) = 0.5$

Then 430's cancel out and we have $P(\frac{-y}{10} \le Z \le \frac{y}{10}) = 0.5$

Then I am confused as to what y values to choose in order to get 0.5. Stats tables are not helping me because I am unsure of which value I am looking for. Can someone lend me a hand?

Answer 1

I think your first statement is incorrect, leading to confusion right from the start.

In terms of $Y \sim Norm(\mu=430, \sigma=10),$ you want to find $b$ such that that $P(Y < b) = .75.$ Then standardize to express the question in terms of standard normal $Z \sim Norm(0,1).$

$$P\left\{Z = \frac{Y-\mu}{\sigma} < \frac{b - 430}{100} \right\} = .75.$$ Then, looking at normal tables you can see that $(b-430)/100 \approx 0.6745,$ which you can solve for $b$. In a similar way, you can find the lower endpoint $a$ of the interval that contains the 'middle half' of the exam scores.

I understand that this is only an outline of the method, but I hope you can take it from there--perhaps by taking excellent Comments into account.

From R software, these values can be obtained without standardizing, and in one statement, as shown below. The answers are not exactly the same because using the software is slightly more accurate than using printed tables. Because exam scores are integers, you probably want to give integer answers.

qnorm(c(.25, .75), 430, 10)
423.255102498039 436.744897501961

It is usually helpful to draw a sketch in solving such problems. The three vertical lines in the figure below break the density curve for $Norm(430,100)$ into four areas, each of 0.25 probability.