Respuesta :
Answer:
0.0228 = 2.28% probability that any shopper selected at random spends more than $80 per week.
88.54% of shoppers are expected to spend between $30 and 80 per week.
Step-by-step explanation:
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.
Normally distributed with a mean of $50 and a standard deviation of $15.
This means that [tex]\mu = 50, \sigma = 15[/tex]
Find the probability that any shopper selected at random spends more than $80 per week?
This is 1 subtracted by the p-value of Z when X = 80. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{80 - 50}{15}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a p-value of 0.9772
1 - 0.9772 = 0.0228
0.0228 = 2.28% probability that any shopper selected at random spends more than $80 per week.
Find the percentage of shoppers who are expected to spend between $30 and 80 per week?
The proportion is the p-value of Z when X = 80 subtracted by the p-value of Z when X = 30.
X = 80
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{80 - 50}{15}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a p-value of 0.9772
X = 30
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{30 - 50}{15}[/tex]
[tex]Z = -1.33[/tex]
[tex]Z = -1.33[/tex] has a p-value of 0.0918
0.9772 - 0.0918 = 0.8854
0.8854*100% = 88.54%
88.54% of shoppers are expected to spend between $30 and 80 per week.
