Question

The amount of corn chips dispensed into a 10-ounce bag by the dispensing machine has been identified at possessing a normal distribution with a mean of 10.5 ounces and a standard deviation of 0.2 ounces (these are the population parameters). Suppose a sample of 100 bags of chips were randomly selected from this dispensing machine. Find the probability that the sample mean weight of these 100 bags exceeded 10.6 ounces. (Hint: think of this in terms of a sampling distribution with sample size = 100)

Answer 1

Solution:

Given that ,

= 10.5

= 0.2

A sample of size n = 100 is taken from this population.

Let be the mean of sample.

The sampling distribution of the is approximately normal with

Mean = = = 10.5

SD = =    = 0.2/​100 = 0.02

Find P( > 10.6)

P( > 10.6)

= P[( - )/ >  (10.6 - )/]

= P[Z > (10.6 - 10.5)/0.02]

= P[Z > 5.00]

= 1 - P[Z < 5.00]

= 1 - 1 ( use z table)

= 0.00000

The probability that the sample mean weight of these 100 bags exceeded 10.6 ounces is zero