Mathematics, 08.04.2021 20:50 cyaransteenberg

Forty families gathered for a fund-raising event. Suppose the individual contribution for each family is normally distributed with a mean and a standard deviation of $105 and $40, respectively. The organizers would call this event a success if the total contributions exceed $4,600. What is the probability that this fund-raising event is a success

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Answer from: KoolCatz

0.0571 = 5.71% probability that this fund-raising event is a success

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 estabilishes 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 organizers would call this event a success if the total contributions exceed $4,600

Forty families, so sample mean above 4600/40 = 115.

Mean and a standard deviation of $105 and $40

This means that $\mu = 105, \sigma = 40$

Sample of 40

This means that $n = 40, s = \frac{40}{\sqrt{40}} = 6.3246$

What is the probability that this fund-raising event is a success?

Sample mean above 115, which is 1 subtracted by the pvalue of Z when X = 115. So

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

By the Central Limit Theorem

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

$Z = \frac{115 - 105}{6.3246}$

Z = 1.58

Z = 1.58 has a pvalue of 0.9429

1 - 0.9429 = 0.0571

0.0571 = 5.71% probability that this fund-raising event is a success

Answer from: Quest

step-by-step explanation:

the solution is in the picture

The circle below is centered at the point (3, 1) and has a radius of length 2. what is the equation?

Answer from: Quest

$22. if larry's profit were to minimize by one for every hour his store were to be closed, we would subtract 30 by 8.

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