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Suppose a random sample of 25 students is selected from a community college where the scores in the final exam (out of 125 points) are normally distributed, with mean equal to 112 and standard deviation equal to 12. What is the mean of the sampling distribution of the sample mean *? What is the standard deviation? Find the probability that * exceeds 116. Find the probability that the sample mean deviates from the population mean = 112 by no more than 4.

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sakhilemdletshesm

Answer:

he probability that the sample mean deviates from the population mean = 112 by no more than 4= 0.9050

Step-by-step explanation:

mean = 112

sd = 12/sqrt(25) = 2.4

z = (x - μ)/σ

z = (116 - 112)/2.4 = 1.67

Therefore,

P(X >= 116) = P(z <= (116 - 112)/2.4)

= P(z >= 1.67)

= 1 - 0.9525

= 0.0475

z = (x - μ)/σ

z1 = (108 - 112)/2.4 = -1.67

z2 = (116 - 112)/2.4 = 1.67

Therefore, we get

P(108 <= X <= 116) = P((116 - 112)/2.4) <= z <= (116 - 112)/2.4)

= P(-1.67 <= z <= 1.67) = P(z <= 1.67) - P(z <= -1.67)

= 0.9525 - 0.0475

= 0.9050

The probability that the sample mean deviates from the population mean of 112 by no more than 4 is 0.9050.

How to calculate the probability?

From the information given, the following can be depicted:

  • Mean = 112
  • Standard deviation = 12/[tex]\sqrt{25}[/tex]  = 2.4
  • z = (x - μ)/σ
  • z = (116 - 112)/2.4 = 1.67

Therefore,

P(X >= 116) = P(z <= (116 - 112)/2.4)

= P(z >= 1.67)

= 1 - 0.9525 = 0.0475

z = (x - μ)/σ

z1 = (108 - 112)/2.4 = -1.67

z2 = (116 - 112)/2.4 = 1.67

Therefore, P(108 <= X <= 116) will be:

= P((116 - 112)/2.4) <= z <= (116 - 112)/2.4)

= P(-1.67 <= z <= 1.67) = P(z <= 1.67) - P(z <= -1.67)

= 0.9525 - 0.0475

= 0.9050

In conclusion, the probability is 0.9050.

Learn more about probability on:

https://brainly.com/question/24756209

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