# STATS HELP!!!?

A particular fruit's weights are normally distributed, with a mean of 338 grams and a standard deviation of 15 grams.

If you pick 16 fruit at random, what is the probability that their mean weight will be between 340 grams and 347 grams

### 2 Answers

- paramvenuLv 710 months ago
Standard deviation of the sampling distribution of x = SEx = Sigma/sqrt n

= 15/sqrt 16

= 3.75

z = (Xbar - Mu)/SEx

z1 = (340 - 338)/3.75 = + 0.53 approximately

z2 = (347 - 338)/3.75 = + 2.40

Required probability = P(340 < Xbar < 347)

= P(0.53 < z < 2.40)

= P(z < 2.40) - P(z < 0.53)

= 0.9918 - 0.7019

= 0.2899

z-score chart showing the area under the standard normal curve on left of z is used...

- az_lenderLv 710 months ago
The standard error of the sampling distribution will be 15/sqrt(16) = 3.75.

The z-scores for the given bounds are

z = 2/3.75 and z = 9/3.75, or

0.5333 < z < 2.4000.

Go to a normal distribution to find what percent of data lie between these two z-scores.

I get around 29%.