Approximately $54 \%$ of mathematics students do their homework on time. In a class of 250 students, what is the mean, varlance, and standard deviation if we assume normality and use the normal distribution as an approximation of the binomial distribution? Answer cholces rounded to the nearest whole number. a.) Mean $=54$ Variance $=8$ Standard Deviation $=62$ b.) Mean $=53$ Varlance $=62$ Standard Deviation $=8$ c.) Mean $=135$ Varlance $=8$ Standard Deviation $=62$ d.) Mean $=135$ Varlance $=62$ Standard Deviation $=8$ SUBMIT MY ANSWER

Step 1 ：The problem is asking for the mean, variance, and standard deviation of the number of students who do their homework on time in a class of 250 students, given that approximately 54% of students do their homework on time.

Step 2 ：The mean of a binomial distribution is given by np, where n is the number of trials (in this case, the number of students) and p is the probability of success (in this case, the probability that a student does their homework on time).

Step 3 ：The variance of a binomial distribution is given by np(1-p), where n is the number of trials, p is the probability of success, and 1-p is the probability of failure.

Step 4 ：The standard deviation is the square root of the variance.

Step 5 ：Given that n=250 and p=0.54, we can calculate the mean, variance, and standard deviation.

Step 6 ：Mean = np = 250 * 0.54 = 135

Step 7 ：Variance = np(1-p) = 250 * 0.54 * (1 - 0.54) = 62.1, which is approximately 62 when rounded to the nearest whole number.

Step 8 ：Standard Deviation = \(\sqrt{Variance}\) = \(\sqrt{62.1}\) = 7.88, which is approximately 8 when rounded to the nearest whole number.

Step 9 ：\(\boxed{\text{Final Answer: The mean is 135, the variance is approximately 62, and the standard deviation is approximately 8. Therefore, the correct answer is (d.) Mean =135, Variance =62, Standard Deviation =8.}}\)