Question

The weight of a beg of pears at the local market averages 8 pounds with a standard deviation of 0.5 pound. The weight of all the bags of pears at the market closely follows a normal distribution. Determined what percentage of bags, to the nearest integer, weighed more than 8.25 pounds.

Answer 1

Mean weight of the bag of pears = u = 8 pounds
Standard deviation = s = 0.5 pounds

We have to find what percentage of bags of pears will weigh more than 8.25 pounds. This can be done using the z score.

We have to convert x = 8.25 to z scores, which will be:

$z= \frac{x-u}{s}= \frac{8.25-8}{0.5}$

z score = 0.5

From the z table, the probability of z score being greater than 0.5 is 0.3085

Therefore, the probability of a bag to weigh more than 8.25 pounds is 0.3085

Thus 0.3085 or 31% (rounded to nearest integer) of bags of pears will have weight more than 8.25 pounds at the local market.