A test consists of 90 mulitiple choice questions, each with five possible answers, only one of which is correct. Find the mean and the standard deviation of the number of correct answers.
mean = np = 90 * 1/5 = ?
(Note: convert any fractions to decimals.)
standard deviation = √npq = √(90 * 1/5 * 4/5) = ?
(Note: q = 1 - p)
I'll let you take it from here.
To find the mean and standard deviation of the number of correct answers in a test with 90 multiple-choice questions, each with five possible answers, we can use the properties of the binomial distribution.
Let's define the random variable X as the number of correct answers. Here, each question has a 1/5 probability of being answered correctly, and since there are 90 questions, the distribution follows a binomial distribution with parameters n = 90 and p = 1/5.
The mean (μ) of a binomial distribution is given by the formula:
μ = n * p
Substituting the values, we have:
μ = 90 * (1/5) = 18
Therefore, the mean number of correct answers is 18.
The standard deviation (σ) of a binomial distribution is given by the formula:
σ = √(n * p * (1 - p))
Substituting the values, we have:
σ = √(90 * (1/5) * (1 - 1/5)) = √(90 * (1/5) * (4/5)) = √(72/5)
Therefore, the standard deviation of the number of correct answers is √(72/5).
To find the mean and standard deviation of the number of correct answers on a test with 90 multiple-choice questions, each with five possible answers, only one of which is correct, we can use probability theory.
The probability of choosing the correct answer for a single multiple-choice question is 1/5, and the probability of choosing an incorrect answer is 4/5.
Let X be a random variable representing the number of correct answers on the test. Since each question on the test is independent of one another, X follows a binomial distribution with parameters n = 90 (number of trials) and p = 1/5 (probability of success).
Mean:
The mean of a binomial distribution is given by the formula μ = n * p. Therefore, the mean number of correct answers on the test is μ = 90 * (1/5) = 18.
Standard Deviation:
The standard deviation of a binomial distribution is given by the formula σ = sqrt(n * p * (1 - p)). Therefore, the standard deviation of the number of correct answers on the test is σ = sqrt(90 * (1/5) * (4/5)) ≈ 6.71.
Thus, the mean number of correct answers on the test is 18, and the standard deviation is approximately 6.71.