Women’s heights are normally distributed with a mean of 162 cm and standard deviation of 16 cm.
a. Define an random variable, X, and describe its full distribution including the mean and variance.
b. What percentage of heights are greater than 180 cm ?
c. What height is at the 90th percentile (i.e. height separating the lower 90% from the upper 10%)?
d. What height is at the 40th percentile?
e. If five students are chosen at random, what is the probability that none of them will be in the 90th percentile or higher?
a. You already have that information given.
b. Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
c, d, e. Use same table.
e. Probability of all events occurring = product of the individual probabilities.
a. The random variable, X, represents the height of women. The distribution of X is a normal distribution, which is also known as a Gaussian distribution or a bell curve. The mean of this distribution is μ = 162 cm, and the standard deviation is σ = 16 cm. The variance, σ^2, is the square of the standard deviation and is equal to 256 cm^2.
b. To find the percentage of heights greater than 180 cm, we need to calculate the area under the normal distribution curve to the right of 180 cm. Since we are dealing with a continuous distribution, we can use the standard normal distribution (z-distribution) to calculate probabilities. We first need to standardize the value 180 cm using the formula z = (x - μ) / σ, where x is the value we want to standardize, μ is the mean, and σ is the standard deviation. In this case, z = (180 - 162) / 16 = 1.125.
Using a standard normal distribution table or a calculator, we can find the probability associated with a z-value of 1.125. Alternatively, we can use a z-table or calculator to find the area to the left of 1.125 and subtract it from 1 to find the area to the right. Either way, we find that the area to the right of 1.125 is approximately 0.1314, or 13.14%. Therefore, approximately 13.14% of the heights are greater than 180 cm.
c. To find the height at the 90th percentile, we need to find the value of X such that 90% of the data falls below that value. We can use the z-score formula again to find the corresponding z-value. The z-value associated with the 90th percentile is found using the z-table or calculator to be approximately 1.282.
Using the z-score formula, we can solve for X:
z = (X - μ) / σ
1.282 = (X - 162) / 16
Multiplying both sides by 16 gives:
20.512 = X - 162
Adding 162 to both sides gives:
X = 182.512
Therefore, the height at the 90th percentile is approximately 182.512 cm.
d. To find the height at the 40th percentile, we follow the same process as before. The z-value associated with the 40th percentile is found using the z-table or calculator to be approximately -0.253.
Using the z-score formula, we can solve for X:
z = (X - μ) / σ
-0.253 = (X - 162) / 16
Multiplying both sides by 16 gives:
-4.048 = X - 162
Adding 162 to both sides gives:
X = 157.952
Therefore, the height at the 40th percentile is approximately 157.952 cm.
e. The probability that none of the five randomly chosen students will be in the 90th percentile or higher can be calculated by finding the probability that each student chosen is below the threshold and multiplying those probabilities together.
The probability that one student is below the threshold is equal to the area under the normal distribution curve to the left of the threshold. Using the z-score formula, we can calculate the z-value associated with the 90th percentile threshold, which is approximately 1.282. The probability of any student being below this threshold is given by the area to the left of this z-value, which can be found using a z-table or calculator.
Let's call this probability P, and since all five students are chosen independently, the probability that none of them will be in the 90th percentile or higher is P^5.