Question

a company is considering a new marketing plan that will cost $600,000. based on answers from a random sample of consumers, the companys marketing analysts have prepared the following table, which shows the probabilities of various payoffs the company could expect from the new plan. the analysts calculate that the mean of the discrete random variable x is $725,000. this is the expected payoff to the company of the marketing plan. the analysts excitedly take this to their boss, who says, wait a minute! this plan is going to cost us $600,000. what about the variance in your estimate? which of the following choices is closest to the standard deviation of x? a.$725,000 b.$125,000 c.$118,000,000,000 d.$344,000

Explanation:

Step1: Recall the formula for variance

The formula for the variance \(\sigma^{2}\) of a discrete random variable \(x\) is \(\sigma^{2}=\sum(x - \mu)^{2}P(x)\), where \(\mu\) is the mean and \(P(x)\) is the probability of \(x\).

Step2: Calculate \((x-\mu)^{2}P(x)\) for each \(x\)

• When \(x = 500000\), \(\mu=725000\), \(P(x)=0.50\)

\((500000 - 725000)^{2}\times0.50=( - 225000)^{2}\times0.50=50625000000\times0.50 = 25312500000\)

• When \(x = 750000\), \(\mu = 725000\), \(P(x)=0.30\)

\((750000 - 725000)^{2}\times0.30=(25000)^{2}\times0.30 = 625000000\times0.30=187500000\)

• When \(x = 1000000\), \(\mu=725000\), \(P(x)=0.15\)

\((1000000 - 725000)^{2}\times0.15=(275000)^{2}\times0.15 = 75625000000\times0.15 = 11343750000\)

• When \(x = 2000000\), \(\mu=725000\), \(P(x)=0.05\)

\((2000000 - 725000)^{2}\times0.05=(1275000)^{2}\times0.05=1625625000000\times0.05 = 81281250000\)

Step3: Calculate the variance \(\sigma^{2}\)

\(\sigma^{2}=25312500000+187500000 + 11343750000+81281250000\)
\(\sigma^{2}=118125000000\)

Step4: Calculate the standard deviation \(\sigma\)

The standard deviation \(\sigma=\sqrt{\sigma^{2}}\), so \(\sigma=\sqrt{118125000000}\approx344000\)

Answer:

D. \(\$344,000\)