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X ~ N(70, 14). Suppose that you form random samples of 25 from this distribution. Let X be the random variable of averages. Let ?X be the random variable of sums. Find the 10th percentile. (Round your answer to two decimal places.)

Sagot :

Answer:

The 10th percentile is 66.42.

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 [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 normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

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

X ~ N(70, 14).

This means that [tex]\mu = 70, \sigma = 14[/tex]

Suppose that you form random samples of 25 from this distribution.

This means that [tex]n = 25, s = \frac{14}{\sqrt{25}} = 2.8[/tex]

Find the 10th percentile.

This is X when Z has a pvalue of 0.1, so X when Z = -1.28.

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

By the Central Limit Theorem

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

[tex]-1.28 = \frac{X - 70}{2.8}[/tex]

[tex]X - 70 = -1.28*2.8[/tex]

[tex]X = 66.42[/tex]

The 10th percentile is 66.42.