Scores on a statistics final in a large class were normally distributed with a mean of 79 and a standard deviation of 12. Use the TI-84 PLUS calculator to answer the following. Round the answers to at least two decimals. (a) Find the 36th percentile of the scores. (b) Find the 70th percentile of the scores. (c) The instructor wants to give an A to the students whose scores were in the top 12% of the class. What is the minimum score needed to get an A

a) The 36th percentile of the scores is of 74.68.

b) The 70th percentile of scores is 85.3.

c) The minimum score needed to get an A is 93.1.

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 79 and a standard deviation of 12.

This means that

(a) Find the 36th percentile of the scores.

This is X when Z has a pvalue of 0.36. So X when Z = -0.36.

The 36th percentile of the scores is of 74.68.

(b) Find the 70th percentile of the scores.

This is X when Z has a pvalue of 0.7, so X when Z = 0.525.

The 70th percentile of scores is 85.3.

(c) The instructor wants to give an A to the students whose scores were in the top 12% of the class. What is the minimum score needed to get an A?

The 100 - 12 = 88th percentile, which is X when Z has a pvalue of 0.88, so X when Z = 1.175.

The minimum score needed to get an A is 93.1.

it would more likely be bald eagle because its has more in the box so about 75%.

step-by-step explanation: