Question

A selective college would like to have an entering class of 1200 students. Because not all students who are offered admission accept, the college admits more than 1200 students. Past experience shows that about 70% of the students will accept. The college decides to admit 1500 students. Assuming that students make their decisions independently, the number who accept has the B(1500, 0.7) distribution. If this number is less than 1200, the college will admit students from its waiting list. Use 4 decimal places in answering each of the following questions. (a) What are the mean and standard deviation of the number X of students who accept? mean: standard deviation: (b) Use the normal approximation to find the probability that at least 1000 students accept. answer: answer (with continuity correction): (c) If the college decides to increase the number of admission offers to 1700, what is the probability that more than 1200 will accept? answer:

Answer 1

The mean and standard deviation of the number of students who accept admission can be calculated using the binomial distribution parameters. The probability that at least 1000 students accept can be found using the normal approximation to the binomial distribution. If the number of admission offers is increased, the probability that more than 1200 students will accept can be calculated using the same method.

Step-by-step explanation:

(a) In this case, X represents the number of students who accept admission. Since the number of students who accept admission has a binomial distribution with parameters n = 1500 and p = 0.7, the mean of X can be calculated as μ = np = 1500 * 0.7 = 1050. The standard deviation of X can be calculated as σ = sqrt(np(1-p)) = sqrt(1500 * 0.7 * (1-0.7)) = 13.2288 (rounded to four decimal places).

(b) To find the probability that at least 1000 students accept, we can use the normal approximation to the binomial distribution. We need to find P(X >= 1000), which is equivalent to 1 - P(X < 1000). Using the normal distribution, we can standardize the value of 1000, find the corresponding z-score, and then use the standard normal table or a calculator to find the probability. The z-score is (1000 - 1050) / 13.2288 = -3.7830 (rounded to four decimal places). Using the standard normal table, we find that the probability is approximately 0.0001 (rounded to four decimal places). Taking into account continuity correction, we can use the normal distribution to approximate the binomial distribution by adding or subtracting 0.5 from the value of X. In this case, we can consider the interval (999.5, 1500.5) and find the probability as P(999.5 < X < 1500.5). Standardizing the values, we find the z-scores as -3.7781 and 2.1919 (rounded to four decimal places). Using the standard normal table, we find the probability as 1 - 0.0001 = 0.9999 (rounded to four decimal places).

(c) If the college decides to increase the number of admission offers to 1700, we need to find the probability that more than 1200 students will accept. Using the same method as in part (b), we can standardize the value of 1200 and find the corresponding z-score. The z-score is (1200 - 1700 * 0.7) / sqrt(1700 * 0.7 * (1-0.7)) = -7.0983 (rounded to four decimal places). Using the standard normal table, we find that the probability is approximately 0 (rounded to four decimal places).

Answer 2

Step-by-step explanation:

Let X be the number of students who decide to get admission

X is Bin (1500, 0.7)

a) Mean = Mean of binomial distribution = np = 1500(0.7) = 1050

Variance = npq = 105 (0.3) = 315

Std dev = 17.75

b) Since np and nq are greater than 5 we can approximate to normal for large n.

X is N(1050, 17.75)

Required prob = prob atleast 1000 students accept.

=
`P(X\geq 1000)\\=P(X\geq 999.5)`(applying continuity correction)

= 0.99778

c) If n increases to 1700 , mean changes to 1190 and std dev to 18.89

`P(X\geq 1200)\\= P(X\geq 1199.5)\\`

=0.3076