Question

the distribution of scores on a recent test closely followed a normal distribution wotb a mean of 22 and a standard deviation of 4 points. what proportion of the stidents scored at least 25 points on this test? what is the 31st percentile of the distribution of test scores

Answer 1

Answer:

1) 22.66%

2) 20

Step-by-step explanation:

The scores of a test are normally distributed.

Mean of the test scores = u = 22

Standard Deviation = $\sigma$ = 4

Part 1) Proportion of students who scored atleast 25 points

Since, the test scores are normally distributed we can use z scores to find this proportion.

We need to find proportion of students with atleast 25 scores. In other words we can write, we have to find:

P(X ≥ 25)

We can convert this value to z score and use z table to find the required proportion.

The formula to calculate the z score is:

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

Using the values, we get:

$z=\frac{25-22}{4}=0.75$

So,

P(X ≥ 25) is equivalent to P(z ≥ 0.75)

Using the z table we can find the probability of z score being greater than or equal to 0.75, which comes out to be 0.2266

Since,

P(X ≥ 25) = P(z ≥ 0.75), we can conclude:

The proportion of students with atleast 25 points on the test is 0.2266 or 22.66%

Part 2) 31st percentile of the test scores

31st percentile means 31%(0.31) of the students have scores less than this value.

This question can also be done using z score. We can find the z score representing the 31st percentile for a normal distribution and then convert that z score to equivalent test score.

Using the z table, the z score for 31st percentile comes out to be:

z = -0.496

Now, we have the z scores, we can use this in the formula to calculate the value of x, the equivalent points on the test scores.

Using the values, we get:

$-0.496=\frac{x-22}{4}\\\\ x=4(-0.496) + 22\\\\ x=20.02\\\\ x \approx 20$

Thus, a test score of 20 represent the 31st percentile of the distribution.