(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
1 answer:
Answer:
The probability is 0.252 ⇒ 3rd answer
Step-by-step 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
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