Answer:
a) 0.073044
b) 0.75033
c) The normal distribution can be used in part (b), even though the sample size does not exceed 30 because initial population size is been distributed normally , therefore, the mean of the samples will be normally distributed regardless of their size(meaning whether the sample size is less than or equal to or exceeds 30, the sample means the mean of the samples will be normally distributed regardless of their size).
Step-by-step explanation:
The overhead reach distances of adult females are normally distributed with a mean of 205.5 and a standard deviation of 8.6 .
a. Find the probability that an individual distance is greater than 218.00 cm.
We solve using z score formula
z = (x-μ)/σ, where
x is the raw score = 218
μ is the population mean = 205.5
σ is the population standard deviation = 8.6
For x > 218
z = 218 - 205.5/8.6
z = 1.45349
Probability value from Z-Table:
P(x<218) = 0.92696
P(x>218) = 1 - P(x<218) = 0.073044
b. Find the probability that the mean for 15 randomly selected distances is greater than 204.00
When = random number of samples is given, we solve using this z score formula
z = (x-μ)/σ/√n
where
x is the raw score = 204
μ is the population mean = 205.5
σ is the population standard deviation = 8.6
n = 15
For x > 204
Hence
z = 204 - 205.5/8.6/√15
z = -0.67552
Probability value from Z-Table:
P(x<204) = 0.24967
P(x>204) = 1 - P(x<204) = 0.75033
c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?
The normal distribution can be used in part (b), even though the sample size does not exceed 30 because initial population size is been distributed normally , therefore, the mean of the samples will be normally distributed regardless of their size(meaning whether the sample size is less than or equal to or exceeds 30, the sample means the mean of the samples will be normally distributed regardless of their size).
to 3. / Your problem should look like: x = -3