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

Answer:

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

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{620 - 500}{80}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

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

Answer 2

Answer:

The z score formula is given by:

$z = \frac{620-500}{80}= 1.5$

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%

Step-by-step 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:

$z=\frac{x-\mu}{\sigma}$

The z score formula is given by:

$z = \frac{620-500}{80}= 1.5$

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%