Question

Weights of the vegetables in a field are normally distributed. From a sample Carl Cornfield determines the mean weight of a box of vegetables to be 172 oz. with a standard deviation of 8 oz. He wonders what percent of the vegetable boxes he has grouped for sale will have a weight less than 164 oz. _
Carl calculates the z-score corresponding to the weight 164 oz. _
Using the table (column .00), Carl sees the area associated with this z-score is 0. Carl rounds this value to the nearest hundredth or 0. _
Now, Carl subtracts 0.50 - 0.34 = 0._ or _%

Answer 1

Answer:

P(X≤164) = 0.16

Step-by-step explanation:

We are given that the weights of vegetables are normally distributed  with mean $\mu$ = 172

Standard deviation, $\sigma$ = 8

We need to calculate P(X≤164) we will use Z score to calculate this,

Z= $\frac{X-\mu}{\sigma}$ = $\frac{164-172}{8}$

Z≤ -1

P(Z≤ -1) = 0.1587 (Using normal distribution table)

Rounding to nearest hundredth = 0.16

Now 0.5-0.34 = 0.16 or 16%