(Binomial to Normal) According to the Gallup poll, 40% of Americans 18 year-old or older stated that they had read at least 6 books within the past year. You conduct a random sample of 200 Americans 18 year- old or older.

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**Title: Converting Binomial Distribution to Normal Approximation**

According to the Gallup poll, 40% of Americans aged 18 and older stated that they had read at least six books within the past year. You conduct a random sample of 200 Americans aged 18 or older. 

**Objective:** Approximate the probability that between 70 and 90, inclusive, have read at least six books within the past year using normal approximation.

**Explanation:** 
This is a classic example of applying the binomial to normal approximation. We begin by identifying the parameters for the binomial distribution: 
- \(n = 200\) (the number of trials)
- \(p = 0.4\) (the probability of success)

Next, we calculate the mean (\(\mu\)) and standard deviation (\(\sigma\)) for the normal distribution approximation: 

- Mean: \(\mu = np = 200 \times 0.4 = 80\)
- Standard Deviation: \(\sigma = \sqrt{np(1-p)} = \sqrt{200 \times 0.4 \times 0.6} \approx 6.93\)

We are interested in the probability that between 70 and 90, inclusive, read at least six books. To find this, we convert the range values to z-scores and use the standard normal distribution table.
Transcribed Image Text:**Title: Converting Binomial Distribution to Normal Approximation** According to the Gallup poll, 40% of Americans aged 18 and older stated that they had read at least six books within the past year. You conduct a random sample of 200 Americans aged 18 or older. **Objective:** Approximate the probability that between 70 and 90, inclusive, have read at least six books within the past year using normal approximation. **Explanation:** This is a classic example of applying the binomial to normal approximation. We begin by identifying the parameters for the binomial distribution: - \(n = 200\) (the number of trials) - \(p = 0.4\) (the probability of success) Next, we calculate the mean (\(\mu\)) and standard deviation (\(\sigma\)) for the normal distribution approximation: - Mean: \(\mu = np = 200 \times 0.4 = 80\) - Standard Deviation: \(\sigma = \sqrt{np(1-p)} = \sqrt{200 \times 0.4 \times 0.6} \approx 6.93\) We are interested in the probability that between 70 and 90, inclusive, read at least six books. To find this, we convert the range values to z-scores and use the standard normal distribution table.
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