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(CO 3) The weights of ice cream cartons are normally distributed with a mean weight of 20.1 ounces and a standard deviation of 0.3 ounces. You randomly select 25 cartons. What is the probability that their mean weight is greater than 20.06 ounces? 0.553 0.748 0.252 0.447

User Santthosh
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1 Answer

2 votes

Answer:

The probability is 0.252 ⇒ 3rd answer

Explanation:

* Lets revise how to find the z-score

- The rule the z-score is z = (M - μ)/σM , where

# M is the sample mean

# μ is the mean

# σM is the standard deviation of the sample mean (standard error)

- The rule of σM = σ/√n , where n is the sample size

* Lets solve the problem

∵ The mean weight is 20.1 ounces

∴ μ = 20.1

∵ The standard deviation is 0.3 ounces

∴ σ = 0.3

∵ You randomly select 25 cartons

∴ n = 25

∵ Their mean weight is greater than 20.06

∴ M = 20.06

- Lets find σM

∵ σM = σ/√n

∴ σM = 0.3/√25 = 0.3/5 = 0.06

- Lets find z-score

∵ z-score = (M - μ)/σM

∴ z-score = (20.06 - 20.1)/0.06 = -0.04/0.06 = - 0.6670

- To find P(M > 20.06) you will asking to find the proportion of area

under the standard normal distribution curve for all z-scores > -0.670

- It can be read from a z-score table by referencing a z-score of -0.670

- Look to the attached table

∴ P(M > 20.06) = 0.2514

* The probability is 0.252

(CO 3) The weights of ice cream cartons are normally distributed with a mean weight-example-1
User Piyush Patil
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5.6k points