Question

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

Answer 1

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

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 = (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 = (\sigma)/(√(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 = (40)/(√(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 = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (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