Answer:
b) 0.0042495
c) 0.048021
Hence, the probability of the mean weight that the 10 apples exceeds 8.2 ounces of this customer wil be higher than the last customer in part b because the number of random samples in part b is greater that that in part c
Step-by-step explanation:
mean of 7.5 ounces and a standard deviation of 1.33 ounce
When random number of samples are given, we solve using z score with the formula
= z = (x-μ)/σ/√n, where
x is the raw score
μ is the population mean
σ is the population standard deviation.
n is the random number of samples
(a) One customer comes to this grocery store and randomly picks 25 apples. What is the mean and standard deviation of the distribution of the mean weight for 25 apples?
(b) What is the probability that the mean weight of 25 apples exceeds 8.2 ounces?
x > 8.2 ounces
= z = (x - μ)/σ/√n
z = (8.2 - 7.5)/1.33/√25
z = (8.2 - 7.5)/1.33/5
z = 2.63158
Probability value from Z-Table:
P(x<8.2) = 0.99575
P(x>8.2) = 1 - P(x<8.2) = 0.0042495
(c) Another customer randomly picks 10 apples. Will the probability of the mean weight that the 10 apples exceeds 8.2 ounces of this customer be higher than the last customer in part b?
For n= 10
x > 8.2 ounces
= z = (x - μ)/σ/√n
z = (8.2 - 7.5)/1.33/√10
z = 1.66436
Probability value from Z-Table:
P(x<8.2) = 0.95198
P(x>8.2) = 1 - P(x<8.2) = 0.048021
Hence, the probability of the mean weight that the 10 apples exceeds 8.2 ounces of this customer wil be higher than the last customer in part b because the number of random samples in part b is greater that that in part c
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