91Ó°ÊÓ

Shopping vehicle and judgment. Refer to the Journal of Marketing Research (December 2011) study of whether you are more likely to choose a vice product (e.g., a candy bar) when your arm is flexed (as when carrying a shopping basket) than when your arm is extended (as when pushing a shopping cart), Exercise 2.85 (p. 112). The study measured choice scores (on a scale of 0 to 100, where higher scores indicate a greater preference for vice options) for consumers shopping under each of the two conditions. Recall that the average choice score for consumers with a flexed arm was 59, while the average for consumers with an extended arm was 43. For both conditions, assume that the standard deviation of the choice scores is 5. Also, assume that both distributions are approximately normally distributed.

a. In the flexed arm condition, what is the probability that a consumer has a choice score of 60 or greater?

b. In the extended arm condition, what is the probability that a consumer has a choice score of 60 or greater?

Step by step solution

01

Given information

Referring to exercise 2.85, the mean of choice of the score for consumers with flexed arms is 59, and the mean of choice of the score for consumers with extended arms is 43,

And the standard deviation for both is 5

Assume that both distributions are approximately normally distributed

02

Calculating for probability

a.

Assume that the choice score for consumers with a flexed arm, i.e., x, follows a normal distribution

The mean and standard deviation of the random variable x is given by,

$$\begin{gathered} \mathrm{μ}=59and\mathrm{σ}=5 \\ x=60 \end{gathered}$$

The z-score is,

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{60-59}{5} \\ =0.2 \\ P(xâ‰\yen 60)=1-P(x<60) \\ =1-P(z<0.2) \\ =1-0.5793 \\ =0.4207 \\ P(xâ‰\yen 60)=0.4207 \end{gathered}$$

Therefore, the probability is, 0.4207

03

Calculating for probability

b.

Assume that the choice score for consumers with an extended arm, i.e., x follows a normal distribution

The mean and standard deviation of the random variable x is given by,

$$\begin{gathered} \mathrm{μ}=43and\mathrm{σ}=5 \\ x=60 \end{gathered}$$

The z-score is,

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{60-43}{5} \\ =3.4 \\ P(xâ‰\yen 60)=1-(x<60) \\ =1-P(z<3.4) \\ =1-0.9996869 \\ =0.0003131 \\ P(xâ‰\yen 60)=0.0003 \end{gathered}$$

Therefore, the probability is, 0.0003

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