Question

10. According to psychologists, IQ's are normally distributed, with a mean of 100 and a standard deviation of 15. a. What percentage of the population has IQ'S between 100 and 130? b. What percentage of the population has IQ's lower than 85?c. What percentage of the population has IQ's higher than 130?

Answer 1

The population of the study is peoples IQ's, this population is known to have a normal distribution with mean μ=100 and standard deviation σ=15.

You have to determine the percentage of people of the population that have determined IQ's, to do so you have to use the standard normal distribution to determine the associated accumulated probabilities within the distribution.

The standard normal distribution is derived from the normal distribution. Considering a random variable X with normal distribution, mean μ and variance δ², the variable Z is determined as follows:

`Z=(X-\mu)/(\sigma)\approx N(0,1)`

The standard normal distribution is tabulated. Any value of any random variable X with normal distribution can be "converted" by subtracting the variable from its mean and dividing it by its standard deviation.

So to calculate each of the asked probabilities, you have to first, "transform" the value of the variable to a value of the standard normal distribution Z, then you use the standard normal tables to reach the corresponding probability.

a. To determine the percentage of the population that has IQ's between 100 and 130, you have to determine the probability of proportion contained between both values, symbolically:

`P(100\leq X\leq130)`

The probability contained in this interval can be calculated by calculating the difference between the accumulated probability until the 130 and the accumulated probability until 100, symbolically:

`P(X\leq130)-P(X\leq100)`

As mentioned before, to determine the associated probabilities you have to work using the standard normal distribution, so the first step is to "translate" both values of IQ into Z-values.

For this, you have to subtract the mean to the IQ value and then subtract it by the standard deviation:

`\begin{gathered} P(Z\leq(130-100)/(15))-P(Z\leq(100-100)/(15)) \\ P(Z\leq2)-P(Z\leq0) \end{gathered}`

Using the Z-table you have to look for the accumulated probability until 2 and the accumulated probability until 0.

`P(Z\leq2)=0.977`
`P(Z\leq0)=0.5`

→ Note that the accumulated probability until zero is 0.5, that is because the center of the distribution is at this value, 50% of the distribution is below it and 50% is above it.

Next, you have to calculate the difference between both probabilities:

`P(Z\leq2)-P(Z\leq0)=0.977-0.5=0.477`

Multiply the result by 100 to express it as a percentage

`0.477\cdot100=47.7`

47.7% of the population has an IQ between 100 and 130

b. You have to determine the percentage of the population that has an IQ below 85. To0 determine this percentage, the first step is to determine the proportion/probability of the population that is below this number, symbolically:

`P(X\leq85)`

To determine this value of probability, first, you have to standardize it(calculate the Z-value), as before, subtract the mean and divide the result by the standard deviation to determine the Z-value

`\begin{gathered} P(Z\leq(X-\mu)/(\sigma)) \\ P(Z\leq(85-100)/(15)) \\ P(Z\leq-1) \end{gathered}`

Next, step is to look for the accumulated probability until -1 in the Z-table:

`P(Z\leq-1)=0.159`

Multiply the probability obtained by 100 to express the result as a percentage

`0.159\cdot100=15.9`

This means that 15.9% of the population has an IQ below 85

c. This time you have to calculate the percentage of the population that has an IQ higher than 130.

So first, let's determine the proportion of the population that has an IQ higher than 130, you can symbolize this as:

`P(X\ge130)`

Next, calculate the corresponding Z-value

`\begin{gathered} P(Z\ge(130-100)/(15)) \\ P(Z\ge2) \end{gathered}`

Note that we need to determine the probability of Z being 2 or higher. In most Z-table, the accumulated probability shown is from minus infinite to the value of Z, symbolically: P(Z≤z₀)

To determine the probability above a number you have to remember that the total probability accumulated in any distribution of probability is 1.

With this in mind, we can determine the probability until "2" and subtract it from the total probability, the result will be the probability above 2:

`P(Z\ge2)=1-P(Z<2)`

We have already looked for the accumulated probability until 2 in item a:

`P(Z<2)=0.977`

So the probability of Z equal to or greater than 2 is equal to 1 minus the probability of Z less than 2:

`1-P(Z<2)=1-0.977=0.023`

Multiply the result by 100 to express it as a percentage:

`0.023\cdot100=2.3`

This result indicates that 2.3% of the population has an IQ greater than 130.