Assume that the population of heights of male college students is approximately normally distributed with mean of 69 inches and standard deviation of 3.75 inches. Show all work.
(A) Find the proportion of male college students whose height is greater than 70 inches.
(B) Find the proportion of male college students whose height is no more than 70 inches.
Is this correct?
a) z=(70-69)/3.75 = .2667
.5+.1026 = .6026
b) 1-.6026 = .374 or 37.4%
any help is appreciated
Your z is correct, but I get 39.5% for (A) and 60.5% for (B)
Clearly there are less than 50% with heights of 70 inches and above if 69 is the mean.
Your calculation is almost correct, but there's a small mistake in part (a) of your calculation.
To find the proportion of male college students whose height is greater than 70 inches, you need to find the area under the normal distribution curve to the right of 70 inches. Here's the correct calculation:
(a) Z = (70 - 69) / 3.75 = 0.2667
To find the area to the right of 70 inches, you need to find the area to the left of -0.2667 (since the Z-tables give probabilities to the left of the Z-score). Looking up the Z-score in a standard normal distribution table, you can find the area to the left of -0.2667, which is approximately 0.3962.
However, we want the proportion of students whose height is greater than 70 inches, so we need to subtract this value from 1 (the total area under the curve):
Proportion = 1 - 0.3962 ≈ 0.6038
So, the proportion of male college students whose height is greater than 70 inches is approximately 0.6038 or 60.38%.
For part (b), you are correct in calculating the proportion of male college students whose height is no more than 70 inches by subtracting the proportion found in part (a) from 1:
(b) Proportion = 1 - 0.6038 = 0.3962
So, the proportion of male college students whose height is no more than 70 inches is approximately 0.3962 or 39.62%.
Overall, your approach is correct, but there was a small error in the calculation for part (a).