A marketing survey is conducted in which students are to taste two different brands of soft drink. Their task is to correctly identify the brand tasted. A random sample of 200 students is taken. Assume that the students have no ability to distinguish between the two brands. Complete​ (a) through​ (c) below.

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Part 1

**a.** What is the probability that the sample will have between​ 50% and 60​% of the identifications​ correct? (Round to four decimal places as​ needed.)

This problem is an example of a **binomial probability distribution**, where:

- There are 200 students (trials),
- The students either correctly or incorrectly identify the brand (two possible outcomes),
- The probability of correctly identifying a brand is 0.5 (assuming no ability to distinguish).

We need to find the probability that the proportion of correct identifications lies between 50% and 60%. This can be expressed as the number of correct identifications falling between 100 and 120 (since \( 50\% \) of 200 is 100, and \( 60\% \) of 200 is 120).

### Step-by-step:

#### 1. Identify the problem as a binomial distribution
- Number of trials \( n = 200 \),
- Probability of success \( p = 0.5 \).

We can use the normal approximation to the binomial distribution since \( n \) is large and \( p \) is close to 0.5.

#### 2. Apply the normal approximation
For large sample sizes, a binomial distribution can be approximated by a normal distribution with:
- Mean \( \mu = np = 200 \times 0.5 = 100 \),
- Standard deviation \( \sigma = \sqrt{np(1-p)} = \sqrt{200 \times 0.5 \times 0.5} = \sqrt{50} \approx 7.0711 \).

#### 3. Convert the range to standard normal variables
We want the probability that the number of correct identifications is between 100 and 120.

Using the continuity correction, we calculate the z-scores for \( X = 99.5 \) and \( X = 120.5 \):
\[
z_1 = \frac{99.5 - 100}{7.0711} = \frac{-0.5}{7.0711} \approx -0.0707
\]
\[
z_2 = \frac{120.5 - 100}{7.0711} = \frac{20.5}{7.0711} \approx 2.898
\]

#### 4. Find the corresponding probabilities
Using the standard normal table, we find:

- The probability for \( z_1 = -0.0707 \) is approximately 0.4717,
- The probability for \( z_2 = 2.898 \) is approximately 0.9981.

#### 5. Compute the final probability
The probability that the sample will have between 50% and 60% correct identifications is:
\[
P(100 \leq X \leq 120) = P(z_1 \leq Z \leq z_2) = 0.9981 - 0.4717 = 0.5264.
\]

Thus, the probability is approximately **0.5264**.

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Do you want more details or have any questions?

Here are some related questions:
1. What is the probability of getting exactly 100 correct identifications?
2. How would the result change if the probability of correctly identifying a brand was 0.6?
3. How do we know when to use the normal approximation for a binomial distribution?
4. What is the probability that more than 130 students correctly identify the brand?
5. How would the standard deviation change if the sample size increased to 500 students?

**Tip**: Always apply the continuity correction when using the normal approximation for a binomial distribution.

A marketing survey is conducted in which students are to taste two different brands of soft drink. Their task is to correctly identify the brand tasted. A random sample of 200 students is taken. Assume that the students have no ability to distinguish between the two brands. What is the probability that the sample will have between 50% and 60% of the identifications correct?

Explore how to calculate the probability of correctly identifying a brand in a marketing survey where students randomly guess between two options. This problem demonstrates the application of binomial distribution and the normal approximation, calculating the probability that 50% to 60% of students correctly identify the brand. Suitable for advanced high school students.

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