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:

what grade level is this? need to know so i can look in my old textbooks!

Answer from: Quest

First, think of what you know. you know that the number t the end of the equation will need to be subtracted because it is the money he lost from buying the lawn mower. and your first two weeks have negative profit, so that tells you that c and d are wrong. now you can plug any of the weeks into a and b to see which one works. i'll start with plugging week 1 into a. $y = 60(1) - 90 \\ y = 60 - 90 \\ y = - 30$ this isn't right because it says your answer for week one should be -90.so now we need to try the same thing with b. $y = 60(1) - 150 \\ y = 60 - 150 \\ y = - 90$ this one got you the right answer, (sometimes if you only check with one of the weeks multiple will work so you need to check with more of the weeks, but it worked out fine here)this means that b is the correct answer.let me know if you need more clarification. i hope this : )

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