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suppose there is a normally distributed population with a mean of 250 and a standard deviation of 50. if xbar is the average of a sample of 36, find P(246

Sagot :

Using the normal distribution and the central limit theorem, it is found that:

[tex]P(246 \leq \bar{x} \leq 260) = 0.5693[/tex]

Normal Probability Distribution

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

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

  • It measures how many standard deviations the measure is from the mean.
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

In this problem:

  • The mean is of [tex]\mu = 250[/tex].
  • The standard deviation is of [tex]\sigma = 50[/tex].
  • A sample of 36 is taken, hence [tex]n = 36, s = \frac{50}{\sqrt{36}} = 8.3333[/tex].

The probability of a sample mean between 246 and 260 is the p-value of Z when X = 260 subtracted by the p-value of Z when X = 246, hence:

X = 260:

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

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{260 - 250}{8.3333}[/tex]

[tex]Z = 1.2[/tex]

[tex]Z = 1.2[/tex] has a p-value of 0.8849.

X = 246:

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{246 - 250}{8.3333}[/tex]

[tex]Z = -0.48[/tex]

[tex]Z = -0.48[/tex] has a p-value of 0.3156.

0.8849 - 0.3156 = 0.5693, hence:

[tex]P(246 \leq \bar{x} \leq 260) = 0.5693[/tex]

To learn more about the normal distribution and the central limit theorem, you can check https://brainly.com/question/24663213

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