Question

An orange juice producer buys oranges from a large orange grove that has one variety of orange. The amount of juice squeezed from these oranges is approximately normally distributed, with a mean of 4.70 ounces and a standard deviation of 0.40 ounce. Suppose that you select a sample of 25 oranges.

a) What is the probability that the sample mean amount of juice will be at least 4.60 ounces?                               _________

b) The probability is 78% that the sample mean amount between what two values around the population mean?       __________

c) The probability is 65% that the sample mean amount of juice will be greater than what value?

Answer 1

(A) Using normalcdf

setting lower = 4.6

upper = 9

mean = 4.7

sigma = 0.4/sqrt(25) = 0.4/5 = 0.08

P(X.4.60) = normalcdf(4.6,9,4.7,0.08)

= 0.8944

(B) probability is 78% that the sample mean amount between two values

so, remaining area = 100-78 = 22%

so, we will have 11% on lower side and 11% on upper side

using z percentile table, z critical corresponding to 11th percentile is -1.227 and it will be 1.227 for upper side(symmetric)

So, lower value = mean +z*sigma

= 4.70 - 1.227*0.08

= 4.6018

and

upper value = mean +z*sigma

= 4.70 + 1.227*0.08

= 4.7982

(C) using invNorm

setting area = 1-0.65 = 0.35

mean = 4.7

sigma = 0.08

we get

Required value = invNorm(0.35,4.7,0.08)

= 4.6692