A distribution of values is normal with a mean of 60 and a standard deviation of 16. From this distribution, you are drawing samples of size 25.

Find the interval containing the middle-most 76% of sample means:

-Enter your answer using interval notation.

-In this context, either inclusive or exclusive intervals would be acceptable.

-Round your numbers to one decimal place.

-Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Respuesta :

Answer:

[tex]\left[56.2, 63.8]\right[/tex]

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means with size n of at least 30 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

In this problem, we have that:

[tex]\mu = 60, \sigma = 16, n = 25, s = \frac{16}{\sqrt{25}} = 3.2[/tex]

Find the interval containing the middle-most 76% of sample means:

50 - 76/2 = 12th percentile to the 50 + 76/2 = 88th percentile.

12th percentile

value of X when Z has a pvalue of 0.12. So X when Z = -1.175

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

By the Central Limit Theorem

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

[tex]-1.175 = \frac{X - 60}{3.2}[/tex]

[tex]X - 60 = -1.175*3.2[/tex]

[tex]X = 56.2[/tex]

88th percentile

value of X when Z has a pvalue of 0.88. So X when Z = 1.175

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

[tex]1.175 = \frac{X - 60}{3.2}[/tex]

[tex]X - 60 = 1.175*3.2[/tex]

[tex]X = 63.8[/tex]

So the answer is:

[tex]\left[56.2, 63.8]\right[/tex]

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