Mathematics, 25.06.2021 01:00 kleathers97

Assume that a population is normally distributed with a mean of 100 and a standard deviation of 15. Would it be unusual for the mean of a sample of 3 to be 115 or more? Why or why not?

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Answer from: gseaton02

|Z| < 2, which means that it would not be unusual for the mean of a sample of 3 to be 115 or more.

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.

If |Z| 2 , the value of X is considered to be unusual.

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.

Assume that a population is normally distributed with a mean of 100 and a standard deviation of 15.

This means that $\mu = 100, \sigma = 15$

Sample of 3

This means that $n = 3, s = \frac{15}{\sqrt{3}}$

Would it be unusual for the mean of a sample of 3 to be 115 or more? Why or why not?

We have to find the z-score.

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

By the Central Limit Theorem

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

$Z = \frac{115 - 100}{\frac{15}{\sqrt{3}}}$

Z = 1.73

|Z| < 2, which means that it would not be unusual for the mean of a sample of 3 to be 115 or more.

Answer from: Quest

correct

step-by-step explanation:

Answer from: Quest

the sum of the lengths of any two sides of a triangle is greater than the length of the third side. if you take the three sides of a triangle and add them in pairs, the sum is greater than (not equal to) the third side. if that is not true, then it is not possible to construct a triangle with the given side lengths.

step-by-step explanation: