The distribution of grades in an introductory finance class is normally distributed, with an expected grade of 68. If the standard deviation of grades is 15, in what range would you expect 68.26 percent of the grades to fall? (Round answers to 2 decimal places, e. g. 15.25. Hint: Think in terms of what the expected highest and lowest scores would be for 68.26% of the students taking the exam.)

The range that you would expect 68.26 percent of the grades to fall is between 53 and 83.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this question:

Middle 68.26% of the grades:

From the

50 - (68.26/2) = 15.87th percentile

To the

50 + (68.26/2) = 84.13rd percentile.

15.87th percentile:

X when Z has a pvalue of 0.1587. So X when Z = -1.

84.13rd percentile:

X when Z has a pvalue of 0.8413. So X when Z = 1.

The range that you would expect 68.26 percent of the grades to fall is between 53 and 83.

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.