Mathematics, 08.07.2021 07:50 ruinmylife

The average height of a current NBA player is 79 inches with a standard deviation of 3.4 inches. A random sample of 35 current NBA players is taken. What is the probability that the mean height of the 35 NBA players will be more than 80 inches?

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Answer from: janessa0502

0.0409 = 4.09% probability that the mean height of the 35 NBA players will be more than 80 inches.

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.

The average height of a current NBA player is 79 inches with a standard deviation of 3.4 inches.

This means that $\mu = 79, \sigma = 3.4$

A random sample of 35 current NBA players is taken.

This means that $n = 35, s = \frac{3.4}{\sqrt{35}}$

What is the probability that the mean height of the 35 NBA players will be more than 80 inches?

This is 1 subtracted by the p-value of Z when X = 80. So

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

By the Central Limit Theorem

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

$Z = \frac{80 - 79}{\frac{3.4}{\sqrt{35}}}$

Z = 1.74

Z = 1.74 has a p-value of 0.9591

1 - 0.9591 = 0.0409

0.0409 = 4.09% probability that the mean height of the 35 NBA players will be more than 80 inches.

Answer from: Quest

step-by-step explanation:

Answer from: Quest

6.5

step-by-step explanation:

you said it was in the question

~courtney

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