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 (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?

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.

=(applying continuity correction)

= 0.99778

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

=0.3076

y = 0.77(x)

step-by-step explanation:

the jacket retains this value after 23% cut or retains 77% of original value. plug original 87 jacket price for x and the remaining value should be y.

answer: since the question is talking about growth you have to use the growth formula which is inital amount (1 + growth)^time

step-by-step explanation:

the intial amount is 1200, the growth is 4.8, and the time is 7. plug those numbers into the equation, also i think you have to change 4.8 into a percent. but, i have forgotton how to do that. if this is a online class go on to the lesson.