Heights of males are norm. dist. w/ u=180cm and o^2=16cm^2 while female heights have u=170cm and o^2=25cm^2.
a)random male and female selected. What is the probability that the male is >5 cm taller? What is the distribution avg of the 2 heights?
b)what is the distribution of the avg of the 2 heights?
Answering as much as possible would be amazing! :)
Proofread your questions before you post them.
Even as the variance, height is a linear dimension and is not squared.
I know it's just a linear dimension, I'm just assuming it's my prof's mistake (as he squared the sd to get variance). I simply copied directly from the paper.
The heights of women are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. The U.S Army require's women's heights of 58 inches and 80 inches. Find the percentage.
Suppose that people's heights (in centimeters) are normally distributed, with a mean of 175 and a standard deviation of 6. We find the heights of 40 people.
a) To find the probability that the male is more than 5 cm taller than the female, we need to calculate the difference in height between the male and female and then find the corresponding probability.
1. Calculate the difference in height between the male and the female:
Male height = 180 cm (mean for males)
Female height = 170 cm (mean for females)
Difference in height = Male height - Female height = 180 cm - 170 cm = 10 cm
2. Determine the standard deviation for the difference in height:
The variance of the sum or difference of two random variables is the sum of their variances.
Variance of male heights = o^2 = 16 cm^2
Variance of female heights = o^2 = 25 cm^2
Variance of the difference in height = Variance of male heights + Variance of female heights = 16 cm^2 + 25 cm^2 = 41 cm^2
Standard deviation of the difference in height = Square root of variance = sqrt(41 cm^2) ≈ 6.403 cm
3. Calculate the z-score for the given height difference:
Z-score = (Observed value - Mean) / Standard deviation
Z-score = (5 cm - 10 cm) / 6.403 cm ≈ -0.782
4. Find the cumulative probability associated with the calculated z-score:
The cumulative probability can be obtained from a standard normal distribution table or using a statistical software. Let's assume it is P(Z > -0.782).
Using a standard normal distribution table or software, the probability is approximately 0.7823. However, since we are interested in the probability that the male is taller, we need to subtract this value from 1:
Probability (male > 5 cm taller) = 1 - 0.7823 ≈ 0.2177
The probability that the male is more than 5 cm taller than the female is approximately 0.2177.
The distribution average of the two heights can be calculated as follows:
Distribution average = (Male mean + Female mean) / 2
Distribution average = (180 cm + 170 cm) / 2 = 350 cm / 2 = 175 cm
b) The distribution of the average of the two heights follows a normal distribution. As both the male and female heights follow normal distributions, the average of the two heights will also be normally distributed. The mean of the distribution of the average of the two heights is equal to the average calculated above, which is 175 cm. The variance of the distribution of the average of the two heights can be calculated using the rule that the variance of the sum or difference of two random variables is the sum of their variances divided by the number of variables being added or subtracted. In this case, since we are calculating the average of two heights, the variance would be the sum of the variances of the male and female heights divided by 2. Therefore, the variance of the distribution of the average of the two heights would be (16 cm^2 + 25 cm^2) / 2 = 41 / 2 = 20.5 cm^2.