Question

A peach farmer has two different varieties (named Variety A and Variety B) of peach trees on her land. Thirty percent of her trees are Variety A peaches and the other 70% are Variety B peaches. The weights of Variety A peaches are normally distributed with a mean of 90 grams and a standard deviation of 8 grams. The weights of Variety B peaches are normally distributed with a mean of 107 grams and a standard deviation of 12 grams. a. What proportion of Variety A peaches weigh between 100 and 110 grams

Answer 1

Answer:

12.03%

Step-by-step explanation:

The z score i used to measure by how many standard deviations the raw score is above or below the mean. It is given by the equation:

$z=\frac{x-\mu}{\sigma}\\ \\where\ \mu=mean,\sigma=standard \ deviation,x=raw\ score$

a) For variety A, the mean = μ = 90, standard deviation = σ = 8

For x = 100 grams:

$z=\frac{x-\mu}{\sigma} =\frac{100-90}{9} =1.11$

For x = 110 grams:

$z=\frac{x-\mu}{\sigma} =\frac{110-90}{9} =2.22$

From the normal distribution table, P(100 < X < 110) = P(1.11 < z < 2.22) = P(z < 2.22) - P(z < 1.11) = 0.9868 - 0.8665 = 0.1203 = 12.03%