Question

Question In an accounting college class, 50% of students receive a B or above on the final exam. 84 students are randomly selected from an accounting class at a local college. Let X be the number of students who received a B or above on the final exam. What normal distribution best approximates X? • Round to one decimal place if entering a decimal answer below.

Answer 1

The normal distribution that best approximates X is N(42, 4.6).

Word problems involving Binomial Distribution.

In the given question, the mean and the standard deviation of the binomial distribution can be estimated by using the formula;

Mean
`(\mu)=n * p`

Mean
`(\mu)=84 * 0.50`

Mean
`(\mu) = 42`

Standard deviation
`(\sigma ) =√(np(1-p))`

Standard deviation
`(\sigma ) =√(84 * 0.50 (1-0.50))`

Standard deviation
`(\sigma ) =√(42 (0.50))`

Standard deviation
`(\sigma ) =√(21.0)`

Standard deviation
`(\sigma ) \simeq 4.6`

Therefore, we can conclude that the normal distribution that best approximates X is N(42, 4.6). It implies that we need about 42 students to receive a B or above with a standard deviation of about 4.6 students.

Answer 2

The number of students who received a B or above on the final exam in the accounting class follows a binomial distribution with a mean of 42 and a standard deviation of 4.6.

Step-by-step explanation:

The number of students who received a B or above on the final exam in an accounting class at a local college follows a binomial distribution since each student has two possible outcomes: receiving a B or above (success) or not receiving a B or above (failure), and the probability of success is constant for each student.

The mean of a binomial distribution is equal to the number of trials (number of students) multiplied by the probability of success. In this case, the mean is 84 * 0.50 = 42.

The standard deviation of a binomial distribution is calculated using the formula √(n * p * (1 - p)), where n is the number of trials and p is the probability of success. For this problem, the standard deviation is √(84 * 0.50 * (1 - 0.50)) = √(21) ≈ 4.6.