Can you help me solve this questions with figures 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. A) What is the probability that the sample will have between 50% and 60% of the identifications correct?
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We have a marketing survey where 200 students taste two different soft drink brands and try to identify them correctly. Since students cannot distinguish between the brands, they are essentially guessing randomly, giving each student a 50% chance of being correct. We need to find the probability that between 50% and 60% of the students will identify the brands correctly.
We have a marketing survey where 200 students taste two different soft drink brands and must identify which one they tasted. Since students cannot distinguish between the brands, each guess is random with 50% chance of being correct. We need to find the probability that between 50% and 60% of identifications are correct.
This problem follows a binomial distribution where X represents the number of correct identifications out of 200 students, with probability 0.5 for each success. The mean is 100 and standard deviation is approximately 7.07. Since both np and n times 1 minus p are greater than 5, we can use the normal approximation to the binomial distribution.
To solve this problem, we convert the percentages to actual counts: 50% of 200 is 100 students, and 60% of 200 is 120 students. We need to find the probability that between 100 and 120 identifications are correct. Using the normal approximation, we standardize these values and find that the probability is approximately 49.77 percent.
Now we convert the percentage range to actual student counts. Fifty percent of 200 students equals 100 students, and sixty percent equals 120 students. Therefore, we need to find the probability that between 100 and 120 students will correctly identify the brands. This corresponds to the shaded area under the normal distribution curve between these two values.
We apply the continuity correction to improve our normal approximation, adjusting our range from 100 to 120 to 99.5 to 120.5. Next, we standardize these values using the Z-score formula. Z1 equals negative 0.071 and Z2 equals 2.90. We now need to find the area under the standard normal curve between these two Z-values, which represents our desired probability.
Now we complete the final calculation. We need to find the probability P of negative 0.071 less than or equal to Z less than or equal to 2.90, which equals phi of 2.90 minus phi of negative 0.071. From the standard normal table, phi of 2.90 is approximately 0.9981 and phi of negative 0.071 is approximately 0.4717. Therefore, our final answer is 0.9981 minus 0.4717 equals 0.5264, or 52.64 percent. This means there is approximately a 52.64 percent chance that between 50 and 60 percent of students will correctly identify the brands by random guessing.