/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Some companies, such as DemandTe... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Some companies, such as DemandTec, have developed software to help retail chains set prices that optimize their profits. An Associated Press story (April 28,2007 ) about this software described a case in which a retail chain sold three similar power drills: one for \(\$ 90\), a better one for \(\$ 120\), and a top-tier one for \(\$ 130\). Software predicted that by selling the middle-priced drill for only \(\$ 110,\) the cheaper drill would seem less a bargain and more people would buy the middle-price drill. a. For the original pricing, suppose \(50 \%\) of sales were for the \(\$ 90\) drill, \(20 \%\) for the \(\$ 120\) drill, and \(30 \%\) for the \(\$ 130\) drill. Construct the probability distribution of \(X=\) selling price for the sale of a drill, and find its mean and interpret, b. For the new pricing, suppose \(30 \%\) of sales were for the \(\$ 90\) drill, \(40 \%\) for the \(\$ 110\) drill, and \(30 \%\) for the \(\$ 130\) drill. Is the mean of the probability distribution of selling price higher with this new pricing strategy? Explain.

Short Answer

Expert verified
Yes, the mean selling price is higher with the new pricing strategy, increasing from \$108 to \$110.

Step by step solution

01

Define Variables and Original Probability Distribution

The selling price variable, \(X\), takes values \(90\), \(120\), and \(130\), with probabilities \(P(X=90) = 0.50\), \(P(X=120) = 0.20\), and \(P(X=130) = 0.30\). This is derived from the original sales proportions.
02

Calculate Mean for Original Pricing

Compute the expected value (mean) of \(X\) for the original distribution: \[ E(X_{original}) = (90 \times 0.50) + (120 \times 0.20) + (130 \times 0.30) \] Calculate this to find: \[ E(X_{original}) = 45 + 24 + 39 = 108 \] Thus, the mean selling price for the original distribution is \$108.
03

Define New Probability Distribution

For the new distribution, the selling price takes values \(90\), \(110\), and \(130\) with probabilities \(P(X=90) = 0.30\), \(P(X=110) = 0.40\), and \(P(X=130) = 0.30\).
04

Calculate Mean for New Pricing Strategy

Compute the expected value (mean) of \(X\) for the new distribution: \[ E(X_{new}) = (90 \times 0.30) + (110 \times 0.40) + (130 \times 0.30) \] Calculate this to find: \[ E(X_{new}) = 27 + 44 + 39 = 110 \] Hence, the mean selling price for the new distribution is \$110.
05

Compare Means and Interpret

Compare the means: \\(108 (original) vs. \\)110 (new). The mean selling price has increased by \$2 with the new pricing strategy. This suggests that the new pricing strategy leads to a higher average revenue per drill sold.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value
Expected value is a crucial concept in probability theory that gives us a measure of the center of a probability distribution. In simple terms, it represents the average outcome if we repeated a random process multiple times. For example, consider the scenario of selling different models of power drills. Each model sells at a specific price with a known probability, which forms our probability distribution. To calculate the expected value of selling price for the drills, we multiply each possible price by its probability and sum these products. This calculation enables businesses to anticipate average sales revenue. Specifically, for the original pricing strategy, we computed the expected value as:- - \(= 90 \times 0.50\)- - \(= 120 \times 0.20\)- - \(= 130 \times 0.30\)Adding these gives us an expected value of \(\$108\). Such insights help retailers make informed decisions about their pricing strategies.
Pricing Strategy
A well-crafted pricing strategy can significantly affect a company's revenue. In the case of the power drills, the objective was to adjust prices to influence sales volume and increase average revenue. For instance, by lowering the price of the mid-tier drill from \(\\(120\) to \(\\)110\), customers perceived the lower-priced drill as less of a bargain, resulting in increased sales of the middle option.The change in pricing not only altered the purchase behavior but also the probability distribution of sales prices. The new distribution showed that the mid-tier drill sold more, shifting sales away from the lowest-priced drill. This shift was intentional and based on understanding consumer psychology—how price differences influence buyers' perceptions and choices.By strategically adjusting prices, businesses can optimize their product positioning in the market. They balance between competing on price and maximizing revenue.
Mean Increase
With changes in the probability distribution due to a better pricing strategy, we expect to see differences in the mean — the average selling price per drill. The mean increase reflects the effectiveness of this strategy in enhancing revenue.In our exercise, altering the price of the mid-tier drill led to a new mean selling price of \(\\(110\), up from the previous \(\\)108\). By calculating the mean in both scenarios:- - Original mean: \(E(X_{original}) = \\(108\)- - New mean: \(E(X_{new}) = \\)110\)The increase of \(\$2\) in the mean selling price indicates a successful strategy shift by boosting overall revenue. Improving the mean selling price by understanding probability distributions validates the change, encouraging further strategic decisions based on robust calculations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Refer to the normal distributions for women's height \((\mu=65, \sigma=3.5)\) and men's height \((\mu=70, \sigma=4.0) .\) A man's height of 75 inches and a woman's height of 70 inches are both 5 inches above their means. Which is relatively taller? Explain why.

In the National Basketball Association, the top free throw shooters usually have probability of about 0.90 of making any given free throw. a. During a game, one such player (Dolph Schayes) shot 10 free throws. Let \(X=\) number of free throws made. What must you assume in order for \(X\) to have a binomial distribution? (Studies have shown that such assumptions are well satisfied for this sport.) b. Specify the values of \(n\) and \(p\) for the binomial distribution of \(X\) in part a. c. Find the probability that he made (i) all 10 free throws and (ii) 9 free throws.

A new roller coaster at an amusement park requires individuals to be at least \(4^{\prime} 8^{\prime \prime}\) (56 inches) tall to ride. It is estimated that the heights of 10-year-old boys are normally distributed with \(\mu=54.5\) inches and \(\sigma=4.5\) inches. a. What proportion of 10 -year-old boys is tall enough to ride the coaster? b. A smaller coaster has a height requirement of 50 inches to ride. What proportion of 10 year-old-boys is tall enough to ride this coaster? c. What proportion of 10 -year-old boys is tall enough to ride the coaster in part b but not tall enough to ride the coaster in part a?

A World Health Organization study (the MONICA project) of health in various countries reported that in Canada, systolic blood pressure readings have a mean of 121 and a standard deviation of \(16 .\) A reading above 140 is considered to be high blood pressure. a. What is the \(z\) -score for a blood pressure reading of \(140 ?\) b. If systolic blood pressure in Canada has a normal distribution, what proportion of Canadians suffers from high blood pressure? c. What proportion of Canadians has systolic blood pressures in the range from 100 to \(140 ?\) d. Find the 90 th percentile of blood pressure readings.

From past experience, a wheat farmer living in Manitoba, Canada finds that his annual profit (in Canadian dollars) is \(\$ 80,000\) if the summer weather is typical, \(\$ 50,000\) if the weather is unusually dry, and \(\$ 20,000\) if there is a severe storm that destroys much of his crop. Weather bureau records indicate that the probability is 0.70 of typical weather, 0.20 of unusually dry weather, and 0.10 of a severe storm. In the next year, let \(X\) be the farmer's profit. a. Construct a table with the probability distribution of \(X\). b. What is the probability that the profit is \(\$ 50,000\) or less? c. Find the mean of the probability distribution of \(X\). Interpret. d. Suppose the farmer buys insurance for \(\$ 3000\) that pays him \(\$ 20,000\) in the event of a severe storm that destroys much of the crop and pays nothing otherwise. Find the probability distribution of his profit.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.