In order to gain additional information about respondents, some marketing researchers have used ultraviolet ink to precode questionnaires that promise confidentiality to respondents. Of 205 randomly selected marketing researchers who participated in an actual survey, 117 said that they disapprove of this practice. Suppose that, before the survey was taken, a marketing manager claimed that at least 65 percent of all marketing researchers would disapprove of the practice.

(a) Assuming that the manager's claim is correct, calculate the probability that 117 or fewer of 205 randomly selected marketing researchers would disapprove of the practice. Use the normal approximation to the binomial. (Round your z answers to 2 decimal places. Round your answer to 4 decimal places. Round the mean and standard deviation to 4 decimal places.)

To solve this problem, we need to use the normal approximation to the binomial distribution.

Here's the setup:

1. **Hypotheses**:
   - The manager claims that \( p = 0.65 \), where \( p \) is the true proportion of researchers who disapprove of the practice.
2. **Given**:
   - Sample size \( n = 205 \)
   - Observed count \( X = 117 \)
   - Hypothesized proportion \( p = 0.65 \)

3. **Goal**:
   - Calculate the probability of observing \( X \leq 117 \) if \( p = 0.65 \).

### Step 1: Mean and Standard Deviation for the Binomial Distribution

For a binomial distribution, the mean \( \mu \) and standard deviation \( \sigma \) are given by:
\[
\mu = n \cdot p
\]
\[
\sigma = \sqrt{n \cdot p \cdot (1 - p)}
\]

Calculating these values:

- \( \mu = 205 \times 0.65 = 133.25 \)
- \( \sigma = \sqrt{205 \times 0.65 \times (1 - 0.65)} \approx 6.7776 \)

### Step 2: Normal Approximation

Using the normal approximation, we can approximate the binomial distribution by a normal distribution with mean \( \mu = 133.25 \) and standard deviation \( \sigma \approx 6.7776 \).

Since we're finding \( P(X \leq 117) \), we need to apply the continuity correction by calculating \( P(X \leq 117.5) \).

### Step 3: Converting to the Standard Normal Distribution

The z-score is calculated by:
\[
z = \frac{X - \mu}{\sigma}
\]

Using \( X = 117.5 \):
\[
z = \frac{117.5 - 133.25}{6.7776} \approx -2.32
\]

### Step 4: Finding the Probability

Now, we look up the z-score of \( -2.32 \) in the standard normal distribution table. This corresponds to a probability of approximately \( 0.0102 \).

### Answer

So, the probability that 117 or fewer out of 205 researchers would disapprove, assuming the manager's claim is correct, is:
\[
\boxed{0.0102}
\]

This low probability suggests that the manager's claim may not be accurate, given that the observed proportion of disapproval is lower than expected under the assumption of \( p = 0.65 \).

Would you like further details on any step, or have additional questions? 

---

**Follow-up Questions:**

1. How would the probability change if the observed count were slightly higher?
2. Why is the continuity correction necessary in normal approximations for discrete data?
3. What are the conditions under which the normal approximation to the binomial distribution is appropriate?
4. How would you test if the manager's claim is statistically valid at a specific significance level?
5. If we surveyed a different sample size, how would it affect the mean and standard deviation in the binomial distribution?

**Tip:** Always ensure sample sizes are large enough when using the normal approximation to the binomial distribution (typically, both \( np \) and \( n(1-p) \) should be at least 5).

This problem explores the use of normal approximation to the binomial distribution to determine the probability that 117 or fewer of 205 marketing researchers disapprove of using ultraviolet ink to precode questionnaires. The analysis involves calculating the mean and standard deviation and finding the corresponding probability, assuming the manager's claim that 65% disapprove is true.

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