Question

Suppose the amount of a popular sport drink in bottles leaving the filling machine has a normal distribution with mean 101.5 milliliters (mL) and standard deviation 1.6. If 36 bottles are randomly selected, find the probability that the mean content is less than 102.1 mL

Answer 1

0.98778

Explanation:

When we are given a random number rod samples, we solve using z score formula

z = (x-μ)/σ/√n, where

x is the raw score = 102.1 mL

μ is the population mean = 101.5 milliliters (mL)

σ is the population standard deviation = 1.6.

n = 36

z = 102.1 - 101.5/1.6/√36

z = 2.25

Probability value from Z-Table:

P(x<102.1) = 0.98778

Hence, If 36 bottles are randomly selected, find the probability that the mean content is less than 102.1 mL is 0.98778