a. Z-score formula
[tex]z=\frac{x-\mu}{\sigma}[/tex]where,
• x: observed value
,• μ: mean
,• σ: standard deviation
Substituting with x = $275, μ = $280, and σ = $20, we get:
[tex]\begin{gathered} z=\frac{275-280}{20} \\ z=-0.25 \end{gathered}[/tex]In terms of the z-score, we need to find
[tex]P(z\ge-0.25)=1-P(z\le-0.25)[/tex]From the table:
[tex]P(z\le-0.25)=0.4013[/tex]Then, the percentage of customers that has daily balances of more than $275 is:
[tex]\begin{gathered} P(z\ge-0.25)=1-0.4013 \\ P(z\ge-0.25)\approx0.6=60\% \end{gathered}[/tex]b. Substituting with x = $243, μ = $280, and σ = $20 into the z-score formula, we get:
[tex]\begin{gathered} z=\frac{243-280}{20} \\ z=-1.85 \end{gathered}[/tex]In terms of the z-score, we need to find:
[tex]P(z\le-1.85)[/tex]From the table, the percentage of customers that has daily balances of less than $243 is:
[tex]P(z\le-1.85)=0.0322=3.22\%[/tex]c. Substituting with x₁ = $241 and x₂ = $301.60, μ = $280, and σ = $20 into the z-score formula, we get:
[tex]\begin{gathered} z_1=\frac{241-280}{20}=-1.95 \\ z_2=\frac{301.60-280}{20}=1.08 \end{gathered}[/tex]In terms of the z-score, we need to find:
[tex]\begin{gathered} P(-1.95\le z\le1.08)=P(-1.95\le z\le0)+P(0\le z\le1.08) \\ P(-1.95\le z\le1.08)=0.5-P(z\le-1.95)+P(0\le z\le1.08) \end{gathered}[/tex]From the first table:
[tex]P(z\le-1.95)=0.0256[/tex]From the second table:
[tex]P(0\le z\le1.08)=0.3529[/tex]Therefore, the percentage of its customers' balances between $241 and $301.60 is:
[tex]\begin{gathered} P(-1.95\le z\le1.08)=0.5-0.0256+0.3529 \\ P(-1.95\le z\le1.08)=0.8273=82.73\% \end{gathered}[/tex]
