A newly hired telemarketer is told he will probably make a sale on about 12% of his phone calls. The first week he called 200 people, but only made 20 sales. Should he suspect he was misled about the true success rate? Explain. Question content area bottom Part 1 Choose the correct answer below. A. Twenty sales is less than 1.0 SDs below the mean. He was probably not probably not misled. B.Twenty sales is more than 3.0 SDs more below the mean. He was probably not misled. C. Twenty sales is more than 0.5 SDs below the mean. He was probably misled. D. Twenty sales is more than 3.0 SDs below the mean. He was probably misled.
The correct answer and explanation is:
Correct Answer: D. Twenty sales is more than 3.0 SDs below the mean. He was probably misled.
Explanation (Approx. 300 Words):
To determine whether the telemarketer was misled about the 12% success rate, we use statistical reasoning. This scenario follows a binomial distribution, since each call is a success (sale) or failure.
Step 1: Define the Binomial Distribution Parameters
- Number of trials (n): 200 calls
- Probability of success (p): 0.12
- Expected number of successes (mean, μ): μ=n⋅p=200⋅0.12=24\mu = n \cdot p = 200 \cdot 0.12 = 24
- Standard deviation (σ) of a binomial distribution: σ=n⋅p⋅(1−p)=200⋅0.12⋅0.88≈21.12≈4.6\sigma = \sqrt{n \cdot p \cdot (1 – p)} = \sqrt{200 \cdot 0.12 \cdot 0.88} \approx \sqrt{21.12} \approx 4.6
Step 2: Calculate the z-score
Now, we compute the z-score to see how far the actual result (20 sales) is from the expected value (24): z=Observed−ExpectedSD=20−244.6≈−44.6≈−0.87z = \frac{\text{Observed} – \text{Expected}}{\text{SD}} = \frac{20 – 24}{4.6} \approx \frac{-4}{4.6} \approx -0.87
Wait — this suggests the observed result is less than 1 standard deviation below the mean, which would contradict option D.
However, here’s the correction: The original mean is 24, but he got 20. That’s a 4-sale deficit.
So again, more accurately: z=20−24200⋅0.12⋅0.88=−421.12≈−44.6≈−0.87z = \frac{20 – 24}{\sqrt{200 \cdot 0.12 \cdot 0.88}} = \frac{-4}{\sqrt{21.12}} \approx \frac{-4}{4.6} \approx -0.87
So, the observed result is only about 0.87 SDs below the mean, not 3.0.
Final Correction:
The correct z-score indicates the deviation is not large enough to conclude he was misled.
Correct Answer (after correction):
✅ A. Twenty sales is less than 1.0 SDs below the mean. He was probably not misled.
Conclusion:
The telemarketer made 20 sales instead of the expected 24. While that is a little low, it’s within one standard deviation of the expected result, which is quite normal in statistical variation. Therefore, there’s no strong evidence he was misled about the 12% success rate.