In North American, female adult heights are approximately normal with a mean of 65 inches and a standard deviation of 3.5 inches.
a.) If one female is selected at Random, what is the probablility that shes has a height 70 inches or higher?
b.) The heights of 50 females were measured at a national collegiate volleyball tournament. The sample mean height was found to be 70 inches. using the population parameters given above, what is the probability of botaining a sample mean height of 70 inches or higher with a random sample of 50 females?
c.)Does the probability you found in (b) make you question the population mean stated for female heights? Justify why you believe this sample mean may not be representative of the population of female heights?
a. 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 that Z score.
b. This is a distribution of means rather than raw scores.
Z = (Sample mean- population mean)/SEm
SEm (standard error of the mean) = SD/√(n-1)
Use same table.
c. Might volleyball players be a biased sample?
To solve these questions, we can use Z-scores and the standard normal distribution. A Z-score measures how many standard deviations a value is from the mean of a distribution.
a) To find the probability that a randomly selected female has a height of 70 inches or higher, we need to find the area to the right of the value on the standard normal distribution.
First, we calculate the Z-score for a height of 70 inches using the formula:
Z = (x - μ) / σ
where x = 70 inches, μ = mean = 65 inches, and σ = standard deviation = 3.5 inches.
Z = (70 - 65) / 3.5 = 1.43
Now, we use a Z-table or a calculator to find the probability associated with this Z-score. The probability is the area to the right of the Z-score. Let's assume the probability is P(Z > 1.43).
b) To find the probability of obtaining a sample mean height of 70 inches or higher with a random sample of 50 females, we use the formula for the standard error of the mean:
SE = σ / sqrt(n)
where σ = standard deviation = 3.5 inches, and n = sample size = 50.
SE = 3.5 / sqrt(50) ≈ 0.4949
Next, we convert the sample mean of 70 inches into a Z-score using the formula:
Z = (x̄ - μ) / SE
where x̄ = sample mean = 70 inches, μ = mean = 65 inches, and SE = standard error = 0.4949.
Z = (70 - 65) / 0.4949 ≈ 10.10
Now, we use a Z-table or a calculator to find the probability associated with this Z-score. The probability is the area to the right of the Z-score. Let's assume the probability is P(Z > 10.10).
c) The probability obtained in part (b) is extremely small and close to zero. This low probability suggests that obtaining a sample mean height of 70 inches or higher with a random sample of 50 females is highly unlikely based on the population parameters given (mean of 65 inches and standard deviation of 3.5 inches).
This result may lead us to question the population mean stated for female heights because the sample mean seems significantly different from it. It is possible that the sample of females at the national collegiate volleyball tournament may not be representative of the population of female heights. Other factors such as self-selection bias or specific characteristics of the tournament participants may contribute to this deviation from the population mean.