SAT : asked on 1fuzzybirdow6e0s
 27.09.2021

A population has a mean of 84 and a standard deviation of 12. A sample of 36 observations will be taken. The probability that the sample mean will be between 80. 54 and 88. 9 is

. 0

Step-by-step answer

09.07.2023, solved by verified expert
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0.00000046997

Explanation:

→ Set up normal distribution without sample size

x ∼ N ( 12 , 84² )A population has a mean of 84 and a standard, №18010536, 27.09.2021 00:06

→ Set up normal distribution with sample size

A population has a mean of 84 and a standard, №18010536, 27.09.2021 00:06

→ State probability to find

p ( 80.54 < x < 88.9 )

→ Use calculator to find the values

Lower = 80.54, Upper = 88.9, σ = √84² ÷ 36 and μ = 12

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SAT
Step-by-step answer
P Answered by Master

0.00000046997

Explanation:

→ Set up normal distribution without sample size

x ∼ N ( 12 , 84² )x

→ Set up normal distribution with sample size

x\alpha N(12,\frac{84^{2} }{36})

→ State probability to find

p ( 80.54 < x < 88.9 )

→ Use calculator to find the values

Lower = 80.54, Upper = 88.9, σ = √84² ÷ 36 and μ = 12

Mathematics
Step-by-step answer
P Answered by PhD

The probability that the sample mean will be between 80.54 and 88.9 is 0.951

Step-by-step explanation:

* Lets revise some definition to solve the problem

- The mean of the distribution of sample means is called M

- The standard deviation of the distribution of sample means is

 called σM

- σM = σ/√n , where σ is the standard deviation and n is the sample size

- z-score = (M - μ)/σM, where μ is the mean of the population  

* Lets solve the problem

∵ The sample size n = 36

∵ The sample mean M is between 80.54 and 88.9

∵ The mean of population μ = 84

∵ The standard deviation σ = 12

- Lets find σM to find z-score  

∵ σM = σ/√n

∴ σM = 12/√36 = 12/6 = 2

- Lets find z-score

∵ z-score = (M - μ)/σM

∴ z-score = (80.54 - 84)/2 = -3.46/2 = -1.73

∴ z-score = (88.9 - 84)/2 = 4.9/2 = 2.45

- Use the normal distribution table to find the probability

∵ P(-1.73 < z < 2.45) = P(2.45) - P(-1.73)

∴ P(-1.73 < z < 2.45) = 0.99286 - 0.04182 = 0.95104

∴ P(-1.73 < z < 2.45) = 0.951

* The probability that the sample mean will be between 80.54 and 88.9

  is 0.951

Mathematics
Step-by-step answer
P Answered by PhD

The probability that the sample mean will be between 80.54 and 88.9 is 0.951

Step-by-step explanation:

* Lets revise some definition to solve the problem

- The mean of the distribution of sample means is called M

- The standard deviation of the distribution of sample means is

 called σM

- σM = σ/√n , where σ is the standard deviation and n is the sample size

- z-score = (M - μ)/σM, where μ is the mean of the population  

* Lets solve the problem

∵ The sample size n = 36

∵ The sample mean M is between 80.54 and 88.9

∵ The mean of population μ = 84

∵ The standard deviation σ = 12

- Lets find σM to find z-score  

∵ σM = σ/√n

∴ σM = 12/√36 = 12/6 = 2

- Lets find z-score

∵ z-score = (M - μ)/σM

∴ z-score = (80.54 - 84)/2 = -3.46/2 = -1.73

∴ z-score = (88.9 - 84)/2 = 4.9/2 = 2.45

- Use the normal distribution table to find the probability

∵ P(-1.73 < z < 2.45) = P(2.45) - P(-1.73)

∴ P(-1.73 < z < 2.45) = 0.99286 - 0.04182 = 0.95104

∴ P(-1.73 < z < 2.45) = 0.951

* The probability that the sample mean will be between 80.54 and 88.9

  is 0.951

SAT
Step-by-step answer
P Answered by PhD
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Step-by-step answer
P Answered by PhD
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Step-by-step answer
P Answered by Master

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Explanation:

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Step-by-step answer
P Answered by Specialist

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Explanation:

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Step-by-step answer
P Answered by Master

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