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

it is part of the rising action that reveals public sentiment toward hester.

explanation:

Answer from: Quest

$x=-2$ or $x=-5$

step-by-step explanation:

we need to solve for 'x' the following equation: $\sqrt{(x+6)} - 4 = x$

we have that:

$\sqrt{(x+6)} - 4 = x$ ⇒ $(x+6) = (x+4)^{2}$

⇒ $x+6 = x^{2} + 8x + 16$ ⇒ $x^{2} + 7x + 10 = 0$

factorizing, we get that:

$x^{2} + 7x + 10 = 0$ ⇒ $(x+2)(x+5)=0$

then, the solutions are: $x=-2$ or $x=-5$

nothing extraneous has been found!

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Forty families gathered for a fund-raising event. Suppose the individual contribution for each famil...