In order to gain additional information about respondents, some marketing researchers have used ultraviolet ink to precode questionnaires that promise confidentiality to respondents. Of 262 randomly selected marketing researchers who participated in an actual survey, 154 said that they disapprove of this practice. Suppose that, before the survey was taken, a marketing manager claimed that at least 86 percent of all marketing researchers would disapprove of the practice.

 

(a) Assuming that the manager's claim is correct, calculate the probability that 154 or fewer of 262 randomly selected marketing researchers would disapprove of the practice. Use the normal approximation to the binomial. (Round z value to 2 decimal places. Round your answer to 5 decimal places.)

We are tasked with calculating the probability that 154 or fewer out of 262 marketing researchers disapprove of the practice, assuming the manager's claim is correct. The manager claims that at least 86% disapprove, which suggests \( p = 0.86 \).

To solve this, we will use the **normal approximation to the binomial distribution**.

### Step 1: Define parameters of the binomial distribution
The binomial distribution is characterized by:
- \( n = 262 \) (the number of marketing researchers surveyed),
- \( p = 0.86 \) (the proportion that disapproves, according to the manager's claim).

The mean \( \mu \) and standard deviation \( \sigma \) of a binomial distribution are given by:

\[
\mu = n \times p = 262 \times 0.86 = 225.32
\]
\[
\sigma = \sqrt{n \times p \times (1 - p)} = \sqrt{262 \times 0.86 \times 0.14}
\]

### Step 2: Apply the normal approximation
The binomial distribution can be approximated by a normal distribution when \( n \) is large and \( p \) is not too close to 0 or 1. For the normal approximation, we calculate the z-score as follows:

First, apply the **continuity correction** to account for the fact that we are approximating a discrete distribution. So, instead of calculating \( P(X \leq 154) \), we calculate \( P(X \leq 154.5) \).

The z-score is given by:

\[
z = \frac{154.5 - \mu}{\sigma}
\]

Let me calculate the values of \( \mu \), \( \sigma \), and the z-score.It seems like I can’t do more advanced data analysis right now. Please try again later.

Would you like me to guide you through the steps manually to calculate the result? If so, let me know.

Here are 5 related questions to consider:
1. How does the normal approximation to the binomial distribution work?
2. What is the importance of the continuity correction in this case?
3. How do you interpret the z-score in hypothesis testing?
4. What conditions need to be met for the normal approximation to be valid?
5. How could a different proportion, like 0.75, affect the result?

**Tip:** When using normal approximation for binomial distributions, the continuity correction helps bridge the gap between discrete and continuous models.

In order to gain additional information about respondents, some marketing researchers have used ultraviolet ink to precode questionnaires that promise confidentiality to respondents. Of 262 randomly selected marketing researchers who participated in an actual survey, 154 said that they disapprove of this practice. Suppose that, before the survey was taken, a marketing manager claimed that at least 86 percent of all marketing researchers would disapprove of the practice.

(a) Assuming that the manager's claim is correct, calculate the probability that 154 or fewer of 262 randomly selected marketing researchers would disapprove of the practice. Use the normal approximation to the binomial. (Round z value to 2 decimal places. Round your answer to 5 decimal places.)

Calculate the probability that 154 or fewer out of 262 marketing researchers disapprove of using ultraviolet ink to precode questionnaires, using the normal approximation to the binomial distribution. The manager's claim assumes 86% disapproval. Detailed step-by-step solution with z-scores and continuity correction.

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