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A salesperson found that there was a 1% chance of a sale from her phone solicitations. Find the probability of getting 5 or more sales for 1000 telephone calls. Round the standard deviation to three decimal places to work the problem.

User Monk L
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Therefore, the probability of getting 5 or more sales for 1000 telephone calls is approximately 0.946, or 94.6%. The rounded standard deviation is 3.15.

To find the probability

we can use the binomial probability formula:

P(X ≥ k) = 1 - P(X < k)

where:

The number of sales is X.

The lowest quantity of sales we desire is k (in this example, 5).

We can determine the likelihood of receiving five or more sales using the following formula, given that the probability of a sale is 1% (0.01) and the probability of not receiving a sale is 99% (0.99):

P(X ≥ 5) = 1 - P(X < 5)

Calculating P(X < 5) using the binomial probability formula is computationally intensive for large values of n and p. Instead, we can use the Normal approximation to the binomial distribution. This approximation is valid when np ≥ 5 and n(1-p) ≥ 5. In this case, np = (1000)(0.01) = 10 and n(1-p) = (1000)(0.99) = 990, which satisfy the conditions for the Normal approximation.

The binomial distribution has np = 10 as its mean and √np(1-p) = √(10)(0.99) = 3.16 as its standard deviation.

We convert the number of sales to z-scores and apply the conventional normal CDF (cumulative distribution function) to determine the chance of having five or more sales. We will modify the algorithm using a continuity correction because the Normal approximation is simply an approximation.

P(Z ≥ (5.5 - 0.5)/3.16) = P(Z ≥ 1.61) ≈ P(X ≥ 5)

We determine that P(Z > 1.61) = 0.946 using a calculator or the usual normal CDF table.

User John Starr Dewar
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