Question

A key metric in the cell phone industry is average revenue per user (ARPU), which represents the average dollar amount that a customer spends per store visit. In 2011, AT&T reported their ARPU as $63.76. Suppose the standard deviation for this population is $22.50. What is the probability that the ARPU will be between $60 and $63 from a random sample of 38 customers?

Answer 1

Answer:

0.266 is the required probability.  

Explanation:

We are given the following information in the question:

Mean, μ = $63.76

Standard Deviation, σ = $22.50

Sample size, n = 38

Standard error due to sampling =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{22.50}{\sqrt{38}} = 3.65$

We have to evaluate:

P(ARPU will be between $60 and $63)

$P(60 \leq x \leq 63) = P(\displaystyle\frac{60 - 63.76}{3.65} \leq z \leq \displaystyle\frac{63-63.76}{3.65}) = P(-1.030 \leq z \leq -0.2082)\\\\= P(z \leq -0.2082) - P(z < -1.030)\\=0.4175 - 0.1515 = 0.266 = 26.6\%$

$P(60 \leq x \leq 63) = 26.6\%$

0.266 is the probability that the ARPU will be between $60 and $63 from a random sample of 38 customers.