In a large class in statistics, the final examination grades have a mean of 67.4 and a standard deviation of 12. Assuming that the distribution of these grades is normal, the number of passes in a class of 180 is: (if you get a number with decimals points, round up to the next whole number)

The number of passes in a class of 180 is 75.

Step-by-step explanation:

The problem does not state, so I will suppose the passing grade is 70.

Problems of normally distributed samples are 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:

Proportion of students who passed:

This is 1 subtracted by the pvalue of Z when X = 70. So

has a pvalue of 0.5871.

1 - 0.5871 = 0.4129

Out of 180:

0.4129*180 = 74.32

Rounding up

The number of passes in a class of 180 is 75.

hello from mrbilldoesmath!

answer:

x - 3

discussion:

x^2 - 9 = (x-3) (x+3)   so

(x^2-9)/ (x + 3) = x-3

you,

mrb