Respuesta :
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 [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]
- 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 [tex]M = n\mu[/tex] and the standard deviation is of [tex]s = \sigma\sqrt{n}[/tex]
In this problem:
- Considering that each pledge is equally as likely, the mean is [tex]\mu = \frac{10 + 50 + 100}{3} = 53.33[/tex]
- The standard deviation is [tex]\sigma = 37[/tex].
- 100 pledges, hence [tex]n = 100, M = 100(53.33) = 5333, s = 37\sqrt{100} = 370[/tex].
The probability that the 100 pledges total more than $5,700 is 1 subtracted by the p-value of Z when X = 5700.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Considering the n instances:
[tex]Z = \frac{X - M}{s}[/tex]
[tex]Z = \frac{5700 - 5333}{370}[/tex]
[tex]Z = 0.99[/tex]
[tex]Z = 0.99[/tex] 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 https://brainly.com/question/24863330
