Question

You solicit 100 pledges for a charitable organization. Each pledge is equally likely to be $10, $50, or $100. You may use the fact that the standard deviation of the three amounts $10, $50 and $100 is $37. What are the chances that the 100 pledges total more than $5,700

Answer 1

Using the normal distribution, it is found that there is a 0.1611 = 16.11% probability that the 100 pledges total more than $5,700.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• It 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, which is the percentile of X.

• For n instances of a normal variable, the mean is $M = n\mu$ and the standard deviation is of $s = \sigma\sqrt{n}$

In this problem:

• Considering that each pledge is equally as likely, the mean is $\mu = \frac{10 + 50 + 100}{3} = 53.33$

• The standard deviation is $\sigma = 37$.

• 100 pledges, hence $n = 100, M = 100(53.33) = 5333, s = 37\sqrt{100} = 370$.

The probability that the 100 pledges total more than $5,700 is 1 subtracted by the p-value of Z when X = 5700.

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

Considering the n instances:

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

$Z = \frac{5700 - 5333}{370}$

$Z = 0.99$

$Z = 0.99$ has a p-value of 0.8389.

1 - 0.8389 = 0.1611

0.1611 = 16.11% probability that the 100 pledges total more than $5,700.

To learn more about the normal distribution, you can take a look at brainly.com/question/24863330