A manufacturer knows that their items have a normally distributed length, with a mean of 10.1 inches, and standard deviation of 0.8 inches.

Question

1. A manufacturer knows that their items have a normally distributed length, with a mean of 10.1 inches, and standard deviation of 0.8 inches.

If 11 items are chosen at random, what is the probability that their mean length is less than 10 inches?

2. A manufacturer knows that their items have a normally distributed lifespan, with a mean of 10 years, and standard deviation of 1.8 years.

If you randomly purchase 24 items, what is the probability that their mean life will be longer than 9 years?

Answer 1

The mean length will be normally distributed with mean = population mean (10.1), standard deviation = population standard deviation / square root of sample size = 0.8 / √ 11

You need p (X<11).

So on a TI-84

2nd VARS

normalcdf

lower = -1E99

upper = 11

mean = 10.1

standard deviation = 0.8 / √11

Hit paste

For (2) similar process. Mean 10, standard deviation 1.8 / √24

p(X>9).

2nd VARS

normalcdf

lower = 9

upper = 1E99

mean = 10

standard deviation = 1.8 / √24

Hit paste