Question

A particular fruit's weights are normally distributed, with a mean of 479 grams and a standard deviation of14 grams.If you pick one fruit at random, what is the probability that it will weigh between 468 grams and 503 grams

Answer 1

Given:

mean = 479 grams

SD = 14 grams

sample fruit between 468 and 503 grams

Find: the probability of a fruit "x" weighing between 468 and 503 grams

Solution:

To be able solve the probability, we will have to convert the weight boundaries to z-value first. The formula is:

$z=\frac{x-\mu}{\sigma}$

where x = sample weight, μ = population mean, and σ = population standard deviation.

Let's start converting the weight boundary x = 468 grams first.

$z=\frac{468-479}{14}=\frac{-11}{14}\approx-0.79$

The z-value of x = 468 grams is -0.79. Using the z-table, the area covered by this z-value to the mean is 0.2852 or 28.52%.

Now, let's convert the other boundary at x = 503 grams.

$z=\frac{503-479}{14}=\frac{24}{14}\approx1.71$

The z-value of x = 503 grams is 1.71. Using the z-table, the area covered from the center mean to 1.71 is 0.4564 or 45.64%.

Graphing the two boundaries in the normal curve, we have:

Hence, the probability of picking a fruit that weighs between 468 grams and 503 grams is (28.52% + 45.64%) 74.16%.