Based on past experience, a telemarketing firm has found that calls to prospective customers take an average of 2.0 minutes, with a standard deviation of 1.5 minutes. The distribution is positively skewed since persons who actually become customers require more of the caller’s time than those who are not home, who simply hang up, or who say they’re not interested. Albert has been given a quota of 220 calls for tomorrow, and he works an 8-hour day. Assuming that this list represents a simple random sample of those persons who could be called, what is the probability that Albert will meet or exceed his quota?

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windyyork windyyork · Answer 1 · 2019-10-17T07:36:56+02:00

Answer: Our required probability would be 0.9641.

Step-by-step explanation:

Since we have given that

Number of hours he works a day = 8

So, Number of minutes he worked in a day = [tex]8\times 60=480\ minutes[/tex]

Number of calls = 220

So, Average [tex]=\bar{x}=\dfrac{480}{220}=2.18\ minutes[/tex]

Standard deviation [tex]=s=\dfrac{\sigma}{\sqrt{n}}=\dfrac{1.5}{\sqrt{220}}=0.10[/tex]

Mean = μ = 2.0 minutes

Standard deviation = σ = 1.5 minutes

Using the normal distribution, we get that

[tex]z=\dfrac{\bar{x}-\mu}{s}\\\\z=\dfrac{2.18-2.0}{0.10}\\\\z=1.8[/tex]

So, the probability that Albert will meet or exceed his quota would be

[tex]P(X\leq 2.18)=P(z\leq 1.8)=0.9641[/tex]

Hence, our required probability would be 0.9641.