Question

a fruit package company produced peaches last summer whose weights were normally distributed with the mean 12 ounces and standard deviation 0.6 ounce. among a sample of 1000 of those peaches about how many could be expected to have any weights of more than 12 ounces?round to the nearest whole number as needed

Answer 1

Let x be the random variable representing the weight of the peaches produced by the company. Since they are normally distributed and the population mean and standard deviation are known, we would apply the formula for normal distribution which is expressed as

z = (x - mean)/standard deviation

z represents the z score

From the informtion given,

mean = 12

standard deviation = 0.6

The probability that the weight of a selected peach is more than 12 ounce is expressed as

P(x > 12)

This is also expressed as

`1\text{ - P(x}\leq12)`

For x = 12,

z = (12 - 12)/0.6 = 0

Looking at the normal distribution table, the probability corresponding to the z score of 0 is 0.5

P(x > 12) = 0.5

`1\text{ - P(x }\leq12)\text{ = 1 - 0.5 = 0.5}`

Since the number of samples is 1000, then the number of peaches expected to have any weights of more than 12 ounces is

0.5 * 1000 = 500

500 peaches are expected to have any weights of more than 12 ounces