# Finding the Probability of a Normal Distribution

The mean IQ scores of 30 primary school students is 108.56 and the Standard deviation is 12.33.

Assume that IQ scores for primary school students that have been kept for 50 years illustrate a normal distribution with a mean of 105 and a standard deviation that agrees with the above.

(i) Write down the value of $y$ where y is 1 standard deviation smaller than the population mean and estimate the probability that the IQ score of a randomly chosen primary school student is greater than $y$.

This is my homework. Should I use sample mean or population mean to obtain the value of $y$? i am confuse which mean to use to calculate $y$. And why?

to determine the if any randomly chosen primary school's IQ score is greater than $y$ is it

$$\frac{\text{IQ score} - 108.56}{12.33} > \frac{y - 105}{12.33 \sqrt{50}}$$

Can help to explain?

I need to know how to work this type of questions

• Both means you mentioned are 105. Where does "108.56" appear? – kennytm May 5 '16 at 5:38
• sorry. The first should be 108.56 . I have corrected it. – Siti May 5 '16 at 5:47

Don't overthink it, just do what it says. It tells you to use the population mean. First write down $y$, that's 1 standard deviation less than the population mean, which is $105-12.33=92.67$
As to why calculate $y$? This is just for the hell of it, to create a number that can be used in the next part of the question so you can learn to work with the concepts!
That is $15.33/12.33=1.243$, so look up 1.243 in normal distribution tables, which gives 0.65. What this means intuitively is that the typical member of your sample has an IQ 1.243 standard deviations above 92.67 and using the normal distribution as a model, this gives a probability of 0.65 that a randomly chosen student from the sample will have an iq greater than that.