Alice earned a score of 940 on an IQ test.The mean test score was 850 with a standard deviation of 100.(Assume that test scores are normally distributed).

Q1a).What proportion of students had a higher score than alice?

Qb)Find the score exceeded by 2.5% of the students.

Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.

The scores on a test are normally distributed with a mean of 140 and a standard deviation of 28, what is the score that is one standard deviation above the mean?

Percentage of scores falling between a z of 1.54 and the mean.

To answer these questions, we need to use the concept of z-scores and the standard normal distribution.

A z-score measures the number of standard deviations a given value is away from the mean. It helps us compare values from different distributions by standardizing them.

Q1a). To find the proportion of students who had a higher score than Alice, we need to calculate the z-score for Alice's score and then find the proportion of scores that fall above that z-score.

First, we calculate the z-score using the formula: z = (X - mean) / standard deviation.

For Alice's score:
X = 940
Mean = 850
Standard Deviation = 100

z = (940 - 850) / 100
z = 0.9
(Note: A positive z-score indicates a value above the mean)

Next, we need to find the proportion of scores that fall above a z-score of 0.9. We can do this by looking up the corresponding value in the z-table or using a calculator.

Using a z-table, we find that the proportion of scores above a z-score of 0.9 is approximately 0.8159.

Therefore, the proportion of students who had a higher score than Alice is approximately 0.8159, or 81.59%.

Qb). To find the score that is exceeded by 2.5% of the students, we need to find the z-score corresponding to this percentile and then convert it back to the actual score.

First, we need to find the z-score that corresponds to the 2.5th percentile. This represents the z-score below which 2.5% of the scores fall.

Using a z-table or calculator, we find that the z-score corresponding to the 2.5th percentile is approximately -1.96.

Next, we can use the formula z = (X - mean) / standard deviation and solve for X.

-1.96 = (X - 850) / 100
-196 = X - 850
X = 654

Therefore, the score exceeded by 2.5% of the students is 654.