The scores of students on a standardized test are normally distributed with a mean of 300 and a standarddeviation of 40. (a) What proportion of scores lie between 220 and 380 points?
(b) What is the probability that a randomly chosen student scores is below 260?
(c) What percent of scores are above 326.8 points?

a) 0.9544 = 95.44% of scores lie between 220 and 380 points.

b) 0.1587 = 15.87% probability that a randomly chosen student scores is below 260.

c) 25.14% of scores are above 326.8 points.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score of a measure X is given by:

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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 300 and a standard deviation of 40.

This means that

(a) What proportion of scores lie between 220 and 380 points?

This is the p-value of Z when X = 380 subtracted by the p-value of Z when X = 220.

X = 380

has a p-value of 0.9772.

X = 220

has a p-value of 0.0228.

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% of scores lie between 220 and 380 points.

(b) What is the probability that a randomly chosen student scores is below 260?

This is the p-value of Z when X = 260. So

has a p-value of 0.1587.

0.1587 = 15.87% probability that a randomly chosen student scores is below 260.

(c) What percent of scores are above 326.8 points?

The proportion is 1 subtracted by the p-value of Z when X = 326.8. So

has a p-value of 0.7486.

1 - 0.7486 = 0.2514

0.2514*100% = 25.14%

25.14% of scores are above 326.8 points.

step-by-step explanation:

hans did not square the 40-unit side; he merely multiplied it by 2.   incorrect.