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denis-greek [22]
3 years ago
14

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

ples of size 25. Find the interval containing the middle-most 76% of sample means.
Mathematics
1 answer:
professor190 [17]3 years ago
3 0

Answer:

The interval containing the middle-most 76% of sample means is between 56.24 and 63.76.

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

A distribution of values is normal with a mean of 60 and a standard deviation of 16.

This means that \mu = 60, \sigma = 16

Samples of size 25:

This means that n = 25, s = \frac{16}{\sqrt{25}} = 3.2

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

Between the 50 - (76/2) = 12th percentile and the 50 + (76/2) = 88th percentile.

12th percentile:

X when Z has a p-value of 0.12, so X when Z = -1.175.

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

-1.175 = \frac{X - 60}{3.2}

X - 60 = -1.175*3.2

X = 56.24

88th percentile:

Z = \frac{X - \mu}{s}

1.175 = \frac{X - 60}{3.2}

X - 60 = 1.175*3.2

X = 63.76

The interval containing the middle-most 76% of sample means is between 56.24 and 63.76.

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