Question

The “last name” effect in purchasing. The Journal of Consumer Research (August 2011) published a study demonstrating the “last name” effect—i.e., the tendency for consumers with last names that begin with a later letter of the alphabet to purchase an item before consumers with last names that begin with earlier letters. To facilitate the analysis, the researchers assigned a number, x, to each consumer based on the first letter of the consumer’s last name. For example, last names beginning with “A” were assigned x = 1; last names beginning with “B” were assigned x = 2; and last names beginning with “Z” were assigned x = 26.

a. If the first letters of consumers’ last names are equally likely, find the probability distribution for x.

b. Find E (x) using the probability distribution, part a. If possible, give a practical interpretation of this value.?

c. Do you believe the probability distribution, part a, is realistic? Explain. How might you go about estimating the true probability distribution for x

Short answer

1. The probability distribution of x is $f\left(x\right)=\frac{1}{n}$ where x = 1,...26
2. $E\left(x\right)=\frac{a+b}{2}$and the interpretation of the value is$$\begin{gathered} E\left(x\right)=\frac{1+26}{2} \\ =13.5 \end{gathered}$$

1. The probability distribution is realistic. Estimating the probability distribution of x is$f\left(x\right)=\frac{1}{n}$

where x = 1,...26.

Step-by-step solution

Given information

Given that researchers assigned a number is x. Here the “last name” effect in purchasing.

Finding the probability distribution for x

a.

The first letters of consumers last names are equally likely. There are equally likely so there follows discrete uniform distribution. Then the probability distribution for x is

$f\left(x\right)=\frac{1}{n}$ where x = 1,...26

Finding E(x) and also define the practical interpretation

b.

The mean value using this distribution to given by

$E\left(x\right)=\frac{a+b}{2}$

The practical interpretation is- Here researchers assigned by numbers. So then the probability distribution of the value is

$$\begin{gathered} E\left(x\right)=\frac{1+26}{2} \\ =13.5 \end{gathered}$$

Probability distribution is realistic and estimating the true probability distribution

c.

Yes the probability distribution is realistic.

The true probability distribution foris defined by discrete uniform distribution. We can also calculate this in step-2.

Thus, the step-2 follows the distribution that will be given by $f\left(x\right)=\frac{1}{n}$where x = 1,...26.