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According to an industry​ report, 26 percent of all households have at least one cell phone.​ Further, of those that do have a cell​ phone, the mean monthly bill is​ $55.90 with a standard deviation equal to​ $9.60. Recently, a random sample of 400 households was selected. Of these​ households, 88 indicated that they had cell phones. The mean bill for these 88 households was​ $57.00. What is the probability of getting 88 or fewer households with cell phones if the numbers provided by the industry report are​ correct?

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Answer:

0.034

Step-by-step explanation:

Given that:

Percentage of households with atleast one cell phone = 26% ; Hence, p = 0.26

Sample size (n) = 400

Using Normal approximation :

Mean(m) = n * p = 400 * 0.26 = 104

Standard deviation (σ) = √(n*p*(1-p))

σ = √(400*0.26*0.74)

σ = 8.7726848

X ≤ 88

Zscore = (x - m) / σ

Zscore = (88 - 104) / 8.7726848

Zscore = - 16 / 8.7726848

Zscore = - 1.8238430

Using the z probability calculator :

P(Z ≤ - 1.8238) = 0.03415

= 0.034

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