Problem 1. A university found that 6% of its students withdraw without completin
ID: 3127085 • Letter: P
Question
Problem 1. A university found that 6% of its students withdraw without completing the introductory statistics course. Assume that 43 students registered for the course and the decisions to withdraw are independent. Write down the name of the distribution and parameter value(s) you used for the below questions (i.e., what table was used).
a. What is the probability that at least 4 will withdraw?
b. What is the probability that at most 5 will withdraw?
c. What is the probability that between 1 and 7 (inclusive) students will withdraw?
d. How many students should we expect to withdraw? Do not round your answer.
Problem 2.. Cars arrive at a car wash randomly and independently; the probability of an arrival is the same for any two time intervals of equal length. The mean arrival rate is 11 cars per hour. Write down the name of the distribution and parameter value(s) you used for each question (i.e., what table was used).
a. What is the probability that at most 20 cars will arrive during the next two hours of operation?
b. What is the probability that more than 22 cars will arrive during the next two hours of operation?
Explanation / Answer
1)6% means = 0.06 probability of withdrawing
a) p(x>=4) = 1 - p(0)+p(1)+p(2)+p(3)
p(0) = 43C0*(0.06)^0(0.94)^43 = 0.069
P(1) = 43C1(0.06)^1*(0.94)^42 = 0.191
P(2) = 43C2*(0.06)^2*(0.94)^41 = 0.257
P(3) = 43C3*(0.06)^3*(0.94)^40 = 0.224
HENCE P(X>=4) = 1 -0.741 = 0.259
B)AT MOST 5 WILL WITHDRAW = P(0)+P(1)+P(2)+P(3)+P(4)+P(5)
P(4) = 43C4(0.06)^4*(0.94)^39 = 0.143
p(5) = 43c5(0.06)^6*(0.94)^38 = 0.100 =
p(x<=5)= 0.741 + 0.243 = 0.984
c)betweeen 1 and 7 = p(1) +......+p(7)
p(6) = 43C6*(0.06)^6*(0.94)^37 = 0.06
P(7) = 43C7*(0.06)^7*(0.94)^36 = 0.05
HENCE THE ANSWER WILL BE = 0.782
D) EX(X) = 0.06*43 = 2.58 = 3
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