Question

33. Suppose that the scores on a statewide standardized test are normally distributed with a mean of 65 and a standard deviation of 5. Estimate the percentage of scores that were(a) between 60 and 70. %(b) above 70. %(c) below 55. %(d) between 60 and 75. %

Answer 1

The scores of the standardized test are normally distributed with a mean, μ, of 75, and a standard deviation, σ, of 4.

To estimate the percentages of each item, you have to work using the standard normal distribution. To standardize each value of the variable you have to subtract the mean and divide the result by the standard deviation following the formula:

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

a) You have to determine the percentage of scores that are between 63 and 87, this can be expressed as follows:

`P(63\leq X\leq87)`

To do so, you have to calculate the difference of the accumulated probability up to 87 and the accumulated probability up to 63

`P(63\leq X\leq87)=P(X\leq87)-P(X\leq63)`

Calculate the corresponding Z values for 63 and 87:

`Z=(63-75)/(4)=-3`
`Z=(87-75)/(4)=3`

Using the standard normal distribution tables you can determine the accumulated probabilities up to each z-value

`P(Z\leq-3)=0.001`
`P(Z\leq3)=0.999`

Now calculate the difference:

`P(X\leq87)-P(X\leq63)\to P(Z\leq3)-P(Z\leq-3)=0.999-0.001=0.998`

The percentage of scores that are between 63 and 87 is 99.8%

b) You have to determine the percentage of scores that are above 83.

`P(X>83)`

Since the distribution tables show accumulated probability, to determine the percentage above 83, you have to work using the complement, that is, subtract from the total probability, what is accumulated until 83:

`P(X>83)=1-P(X\leq83)`

Calculate the Z-value corresponding to X=83

`Z=(83-75)/(4)=2`

The accumulated probability up to Z=2 is:

`P(Z\leq2)=0.977`

Next, calculate the asked probability:

`1-P(X\leq83)\to1-P(Z\leq2)=1-0.977=0.023`

Above 83 you will find 2.3% of the scores.

c) To find the percentage of scores below 71, you have to find the accumulated probability up to the given score:

`P(X\leq71)`

Calculate the corresponding Z-value:

`\begin{gathered} P(Z\leq(71-75)/(4)) \\ P(Z\leq-1)=0.159 \end{gathered}`

The percentage of scores below 71 is 15.9%

d) To determine the percentage of scores between 67 and 79 you have to follow the same steps as in item a:

`P(67\leq X\leq79)=P(X\leq79)-P(X\leq67)`

Find both Z-values and their corresponding probabilities:

`P(Z\leq(79-75)/(4))=P(Z\leq1)=0.841`
`P(Z\leq(67-75)/(4))=P(Z\leq-2)=0.023`

Finally, calculate the difference between both values:

`P(X\leq79)-P(X\leq67)\to P(Z\leq1)-P(Z\leq-2)=0.841-0.023=0.818`

81.8% of the scores are between 67 and 79.