Respuesta :
Answer:
The mean of the number of people ordering the program is 22,046.5 and the standard deviation is 4,651.16.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
Past history shows that 33.00% of the time fewer than 20,000 people order the program
This means that X = 20000 has a pvalue of 0.33. So when X = 20000, Z = -0.44.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.44 = \frac{20000 - \mu}{\sigma}[/tex]
[tex]20000 - \mu = -0.44\sigma[/tex]
[tex]\mu = 20000 + 0.44\sigma[/tex]
Only ten percent of the time do more than 28,000 people order the program.
This means that X = 28000 has a pvalue of 1-0.1 = 0.9. So when X = 28000, Z = 1.28.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{28000 - \mu}{\sigma}[/tex]
[tex]28000 - \mu = 1.28\sigma[/tex]
[tex]\mu = 28000 - 1.28\sigma[/tex]
Also
[tex]\mu = 20000 + 0.44\sigma[/tex]
So
[tex]20000 + 0.44\sigma = 28000 - 1.28\sigma[/tex]
[tex]1.72\sigma = 8000[/tex]
[tex]\sigma = \frac{8000}{1.72}[/tex]
[tex]\sigma = 4651.16[/tex]
[tex]\mu = 20000 + 0.44\sigma = 20000 + 0.44*4651.16 = 22046.5[/tex]
The mean of the number of people ordering the program is 22,046.5 and the standard deviation is 4,651.16.
