Question

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According to a recent survey the mean monthly cell phone for Canadians is $68.72. Suppose monthly cell phone bills are normally distributed with a standard deviation of $8.46. What is the probability that a randomly selected cell phone bill is between $60 and $70? Q 0.55% 0 0.1515 0 0.4081 0 0.4042

Answer & Explanation

Solved by verified expert

answer = 0.4081

Step-by-step explanation

we have that $μ=68.72$ and $σ=8.46$

$z=σx−μ $ hence

$P(60<x<70)=P(8.4660−68.72 <z<8.4670−68.72 )$

=P(-1.03< z < 0.15)

=P(z <0.15) - P(z < -1.03)

Now from z table

=0.5596 - 0.1515

= 0.4081