Question

Test scores are normally distributed with a mean of 500. Convert the given score to a z-score, using the given standard deviation. Then find the percentage of students

who score below 620, if the standard deviation is 80.​

Answer 1

The z score formula is given by:

`z = (620-500)/(80)= 1.5`

And if we use the normal standard distribution table or excel we got:

`P(z<1.5)=0.933`

And the percentage would be 93.3%

Explanation:

Let X the random variable that represent the test scores of a population, and for this case we know the distribution for X is given by:

`X \sim N(500,80)`

Where
`\mu=500` and
`\sigma=80`

We are interested on this probability

`P(X<620)`

The z score formula is given by:

`z=(x-\mu)/(\sigma)`

The z score formula is given by:

`z = (620-500)/(80)= 1.5`

And if we use the normal standard distribution table or excel we got:

`P(z<1.5)=0.933`

And the percentage would be 93.3%

Answer 2

The percentage of students who scored below 620 is 93.32%.

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

`\mu = 500, \sigma = 80`

Percentage of students who scored below 620:

This is the pvalue of Z when X = 620. So

`Z = (X - \mu)/(\sigma)`

`Z = (620 - 500)/(80)`

`Z = 1.5`

`Z = 1.5` has a pvalue of 0.9332

The percentage of students who scored below 620 is 93.32%.