Assuming that the heights of college women are normally distributed with mean 62 inches and standard deviation 3.5 inches, what percentage of women are between 58.5 inches and 72.5 inches?
34.1%
84.0%
15.7% my answer was 13.6
13.6% is this correct
97.6%
Z = (score - mean)/SD
Z = (58.5-62)/ 3.5 = -1
Z = (72.5-62)/3.5 = 3
It goes from one standard deviation below the mean to 3 above.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to those Z scores.
my answer is equals 15.7 is this correct
Assuming that the heights of college women are normally distributed with mean 62 inches and standard deviation 3.5 inches, what percentage of women are between 58.5 inches and 72.5 inches?
a. 13.6%
b. 84.0%
c. 97.6%
d. 34.1%
e. 15.7%
Well, to calculate the percentage of women between 58.5 inches and 72.5 inches, we need to find the area under the normal distribution curve within that range.
First, let's find the z-scores for both values:
For 58.5 inches: (58.5 - 62) / 3.5 ≈ -0.8571
For 72.5 inches: (72.5 - 62) / 3.5 ≈ 2.9286
Now, we can use a Z-table or a calculator to find the area under the normal curve between these z-scores.
Using the Z-table, we find that the area for -0.8571 is approximately 0.1967, and the area for 2.9286 is approximately 0.9981.
To find the percentage between 58.5 and 72.5 inches, we subtract the area for -0.8571 from the area for 2.9286:
0.9981 - 0.1967 = 0.8014
So, approximately 80.14% of college women are between 58.5 inches and 72.5 inches.
Therefore, none of the answer choices provided (34.1%, 84.0%, 15.7%, 13.6%, or 97.6%) is correct.
To find the percentage of women between 58.5 and 72.5 inches, we can use the standard normal distribution table.
First, we need to standardize the values using the formula:
z = (x - μ) / σ
where z is the standardized score, x is the given height, μ is the mean, and σ is the standard deviation.
For 58.5 inches:
z1 = (58.5 - 62) / 3.5
For 72.5 inches:
z2 = (72.5 - 62) / 3.5
Using a standard normal distribution table or a calculator, we can find the corresponding values for z1 and z2.
Assuming a standard normal distribution, the percentage of women between 58.5 and 72.5 inches can be found by calculating the area under the curve between these two z-scores.
To get the percentage, subtract the cumulative probability for z1 from the cumulative probability for z2.
Let's go through the calculations:
z1 = (58.5 - 62) / 3.5 = -1
z2 = (72.5 - 62) / 3.5 = 3
Using the standard normal distribution table, we find the following cumulative probabilities:
P(z < -1) = 0.1587
P(z < 3) = 0.9987
To find the percentage between z1 and z2, we subtract the lower cumulative probability from the higher cumulative probability:
P(-1 < z < 3) = P(z < 3) - P(z < -1)
= 0.9987 - 0.1587
= 0.84
Therefore, the correct answer is 84.0%.