Answer:
69.15% probability that a randomly selected customer spends less than $105 at this store
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.
In this problem, we have that:
[tex]\mu = 100, \sigma = 10[/tex]
What is the probability that a randomly selected customer spends less than $105 at this store?
This is the pvalue of Z when X = 105. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{105 - 100}{10}[/tex]
[tex]Z = 0.5[/tex]
[tex]Z = 0.5[/tex] has a pvalue of 0.6915
69.15% probability that a randomly selected customer spends less than $105 at this store
Answer:
The probability that a randomly selected customer spends less than $105 at this store is 0.6915 or 69.15%
Step-by-step explanation:
1. Let's review the information given to us to answer the question correctly:
Mean of the purchase amounts by customers entering a popular retail store = $ 100
Standard deviation of the purchase amounts by customers entering a popular retail store = $ 10
2. What is the probability that a randomly selected customer spends less than $105 at this store?
For calculating the probability, we need to find the z-score first, this way:
X = 105
z-score = (X - μ)/σ
Replacing with the values we have:
z-score = 105 - 100/10
z-score = 5/10 = 0.5
Now, let's calculate the probability, using the z-table:
p (z = 0.5) = 0.6915
The probability that a randomly selected customer spends less than $105 at this store is 0.6915 or 69.15%
