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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 _%

Respuesta :

Answer:

P(X≤164) = 0.16

Step-by-step explanation:

We are given that the weights of vegetables are normally distributed  with mean [tex]\mu[/tex] = 172

Standard deviation, [tex]\sigma[/tex] = 8

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

Z= [tex]\frac{X-\mu}{\sigma}[/tex] = [tex]\frac{164-172}{8}[/tex]

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%

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