Chapter 6: Problem 89
a. Find a \(z_{0}\) such that \(P\left(z>z_{0}\right)=.9750 .\) b. Find a \(z_{0}\) such that \(P\left(z>z_{0}\right)=.3594\).
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Chapter 6: Problem 89
a. Find a \(z_{0}\) such that \(P\left(z>z_{0}\right)=.9750 .\) b. Find a \(z_{0}\) such that \(P\left(z>z_{0}\right)=.3594\).
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A normal random variable \(x\) has mean \(\mu=10\) and standard deviation
\(\sigma=2\). Find the probabilities of these \(x\) -values:
a. \(x>13.5\)
b. \(x<8.2\)
c. \(9.4
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In Exercise 6.28 , we suggested that the IRS assign auditing rates per state by randomly selecting 50 auditing percentages from a normal distribution with a mean equal to \(1.55 \%\) and a standard deviation of \(.45 \%\) a. What is the probability that a particular state would have more than \(2 \%\) of its tax returns audited? b. What is the expected value of \(x\), the number of states that will have more than \(2 \%\) of their income tax returns audited? c. Is it likely that as many as 15 of the 50 states will have more than \(2 \%\) of their income tax returns audited?
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Students very often ask their professors whether they will be "curving the grades." The traditional interpretation of "curving grades" required that the grades have a normal distribution, and that the grades will be assigned in these proportions: $$ \begin{array}{l|lllll} \text { Letter Grade } & \mathrm{A} & \mathrm{B} & \mathrm{C} & \mathrm{D} & \mathrm{F} \\ \hline \text { Proportion of Students } & 10 \% & 20 \% & 40 \% & 20 \% & 10 \% \end{array} $$ a. If the average "C" grade is centered at the average grade for all students, and if we assume that the grades are normally distributed, how many standard deviations on either side of the mean will constitute the "C" grades? b. How many deviations on either side of the mean will be the cutoff points for the "B" and "D" grades?
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