The CEO of a food store has determined that the weekly demand for a popular type of granola is a normally distributed random variable with mean 95 pounds and standard deviation 10 pounds. If the demand for a given week falls within the lowest 3.5% of all possible values for the weekly demand, the price of the granola will be reduced for the following week. Calculate the value in pounds (lbs) for the weekly demand below which the manager will have to reduce the price.
Using R to solve a problem about a normal distribution
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$\begingroup$
probability
statistics
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1Perhaps r means R, as in https://cran.r-project.org/ – 2017-02-09
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0Good point. I use R, but I was distracted by the lower-case r. – 2017-02-09
1 Answers
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Using tables: $$ 0.035 = P(Z \le -1.82) = P\left(\frac{X - \mu}{\sigma} = \frac{X-95}{10} = -1.82 \right).$$ Solve for $X.$
Using R (following @awkward's Comment): Interpret the following R statements (one for the solution, two for verification). No 'standardization' is necessary:
qnorm(.035, 95, 10)
## 76.88089
pnorm(77, 95, 10)
## 0.03593032
pnorm(76, 95, 10)
## 0.02871656
Sketch: The area under the density curve to the left of the vertical red line is about 0.035.
curve(dnorm(x, 95, 10), 60, 130, lwd=2, col="blue", ylab="PDF",
main="Density of NORM(95, 10)")
abline(h = 0, col="green2")
abline(v = 77.88, lwd=2, col="red")
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0If you don't know the distinctions among `dnorm`, `pnorm`, `qnorm`, and `rnorm`, look at the help screen available from `> ? pnorm`. – 2017-02-09