library(ggplot2)

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Problem 1: Suppose family incomes in a town are normally distributed with a mean of \(Php25000\) and a standard deviation of \(Php6000\) per month. What is the probability that a family has an income between \(Php14000\) and \(Php32500\)? (5 pts.)

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Solution:

# Solve for z-scores:
z1 <- (14000-25000)/6000
z1
  [1] -1.833333
z2 <- (32500-25000)/6000
z2
  [1] 1.25

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z <- seq(from = -4, to = +4, length.out = 1000)
normprob <- data.frame(x = z, y = dnorm(z, mean = 0, sd = 1))
stdnormal <- ggplot(normprob, aes(x,y)) + geom_line() + labs(title = "Standard Normal Distribution", y = "Density", x = "Z")
stdnormal + geom_ribbon(data = subset(normprob, z > z1 & z < z2), aes(ymax = y), ymin = 0, fill = "blue", alpha = 0.5)

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To compute for the required probability:

pnorm(z2) - pnorm(z1)
  [1] 0.8609737

\(~\) The probability that a family has an income between \(Php14000\) and \(Php32500\) is \(0.8610\).

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Problem 2: The amount of time required per individual at a bank teller’s window has been found to be approximately normally distributed with \(\mu\) = \(130\) sec and \(\sigma\) = \(45\) sec. What is the probability that a randomly selected individual will

a.) require less than \(100 sec\) to complete a transaction? (3 pts.) \(~\)

Solution:

# Compute for corresponding z-score:
z1 <- (100-130)/45
z1
  [1] -0.6666667

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z <- seq(from = -4, to = +4, length.out = 1000)
normprob <- data.frame(x = z, y = dnorm(z, mean = 0, sd = 1))
stdnormal <- ggplot(normprob, aes(x,y)) + geom_line() + labs(title = "Standard Normal Distribution", y = "Density", x = "Z")
stdnormal + geom_ribbon(data = subset(normprob, z < z1), aes(ymax = y), ymin = 0, fill = "blue", alpha = 0.5)

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To compute for the required probability:

pnorm(z1)
  [1] 0.2524925

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The probability that a randomly selected individual will require less than \(100 sec\) to complete a transaction is 0.2525.

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b.) spend between \(2.0\) and \(3.0 min\) at the teller’s window? (5 pts.) \(~\)

Solution:

# Compute for corresponding z-scores:
z1 <- (120-130)/45
z1
  [1] -0.2222222
z2 <- (180-130)/45
z2
  [1] 1.111111

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z <- seq(from = -4, to = +4, length.out = 1000)
normprob <- data.frame(x = z, y = dnorm(z, mean = 0, sd = 1))
stdnormal <- ggplot(normprob, aes(x,y)) + geom_line() + labs(title = "Standard Normal Distribution", y = "Density", x = "Z")
stdnormal + geom_ribbon(data = subset(normprob, z > z1 & z < z2), aes(ymax = y), ymin = 0, fill = "blue", alpha = 0.5)

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To compute for the required probability:

pnorm(z2) - pnorm(z1)
  [1] 0.4546693

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The probability that a randomly selected individual will spend between \(2.0\) and \(3.0 min\) at the teller’s window is \(0.4547\).

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Problem 3: The waiters in a restaurant receive an average tip of \(\$20\) per table with a standard deviation of \(\$5\). The amounts of tips are normally distributed, and a waiter feels that he has provided excellent service if the tip is more than \(\$25\). What is the probability that a waiter has provided excellent service to a table? (4 pts.)

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Solution:

# Compute for the corresponding z-score:
z1 <- (25-20)/5
z1
  [1] 1

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z <- seq(from = -4, to = +4, length.out = 1000)
normprob <- data.frame(x = z, y = dnorm(z, mean = 0, sd = 1))
stdnormal <- ggplot(normprob, aes(x,y)) + geom_line() + labs(title = "Standard Normal Distribution", y = "Density", x = "Z")
stdnormal + geom_ribbon(data = subset(normprob, z > z1), aes(ymax = y), ymin = 0, fill = "blue", alpha = 0.5)

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To compute for the required probability:

pnorm(z1, lower.tail = F)
  [1] 0.1586553

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The probability that a waiter has provided excellent service to a table is \(0.1587\).

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Problem 4: An appliance center guarantees their washing machine for a period of 1 year. The washing machine life is normally distributed with a mean of \(3.7\) years and a standard deviation of \(1.8\) years. What is the probability that a washing machine will break down during the guarantee period of \(1\) year? (3 pts.)

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Solution:

# Compute for the corresponding z-score:
z1 <- (1-3.7)/1.8
z1
  [1] -1.5

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z <- seq(from = -4, to = +4, length.out = 1000)
normprob <- data.frame(x = z, y = dnorm(z, mean = 0, sd = 1))
stdnormal <- ggplot(normprob, aes(x,y)) + geom_line() + labs(title = "Standard Normal Distribution", y = "Density", x = "Z")
stdnormal + geom_ribbon(data = subset(normprob, z < z1), aes(ymax = y), ymin = 0, fill = "blue", alpha = 0.5)

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Computing for the required probability:

pnorm(z1)
  [1] 0.0668072

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The probability that a washing machine will break down during the guarantee period of \(1\) year is \(0.0668\).

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Problem 5: A company pays its employees an average wage of \(\$22.50\) an hour with a standard deviation of \(\$2.30\). If the wages are approximately normally distributed and paid to the nearest centavos,
a.) what percentage of the workers receive wages between \(\$20\) and \(\$23.50\) (5 pts.)

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Solution:

# Compute for corresponding z-scores:
z1 <- (20-22.50)/2.30
z1
  [1] -1.086957
z2 <- (23.50-22.50)/2.30
z2
  [1] 0.4347826

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z <- seq(from = -4, to = +4, length.out = 1000)
normprob <- data.frame(x = z, y = dnorm(z, mean = 0, sd = 1))
stdnormal <- ggplot(normprob, aes(x,y)) + geom_line() + labs(title = "Standard Normal Distribution", y = "Density", x = "Z")
stdnormal + geom_ribbon(data = subset(normprob, z > z1 & z < z2), aes(ymax = y), ymin = 0, fill = "blue", alpha = 0.5)

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Computing for the required proportion:

pnorm(z2)-pnorm(z1)
  [1] 0.5296119

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\(52.96\%\) of the workers receive wages between \(\$20\) and \(\$23.50\) an hour.

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b.) the highest \(5\%\) of the employees have hourly wages that are greater than what amount? (5 pts.)

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Solution:

z <- seq(from = -4, to = +4, length.out = 1000)
normprob <- data.frame(x = z, y = dnorm(z, mean = 0, sd = 1))
stdnormal <- ggplot(normprob, aes(x,y)) + geom_line() + labs(title = "Standard Normal Distribution", y = "Density", x = "Z")
stdnormal + geom_ribbon(data = subset(normprob, z > qnorm(0.05, lower.tail = F)), aes(ymax = y), ymin = 0, fill = "blue", alpha = 0.5)

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The z-value with the area to its right equal to \(0.05\) is:

qnorm(0.05, lower.tail = F)
## [1] 1.644854

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Now, to determine the required hourly wage value corresponding to this z score value:

x <- qnorm(0.05, lower.tail = F, mean = 22.50, sd = 2.30)
x
## [1] 26.28316

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The highest \(5\%\) of the employees have hourly wages that are greater than \(\$26.28\).