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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

Sagot :

Answer:

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

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 [tex]\mu = 105, \sigma = 40[/tex]

Sample of 40

This means that [tex]n = 40, s = \frac{40}{\sqrt{40}} = 6.3246[/tex]

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{115 - 105}{6.3246}[/tex]

[tex]Z = 1.58[/tex]

[tex]Z = 1.58[/tex] has a pvalue of 0.9429

1 - 0.9429 = 0.0571

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

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