University Statistics and Probability Homework: Complete Solutions

This document presents solutions to a statistics and probability homework assignment, addressing various problems related to statistical concepts. The solutions cover calculating Z-values, determining means, and working with probabilities. The assignment includes problems on binomial probability, conditional probability, and independence of events. Solutions are provided for questions involving expected values, disjoint events, and the application of standard deviations. The document offers detailed explanations and calculations for each problem, providing a comprehensive resource for students studying statistics and probability. Topics include finding probabilities using Z-tables, determining expected values in different scenarios, and understanding concepts like disjoint and independent events. Overall, the assignment provides a strong foundation in core statistical concepts and problem-solving techniques.

Question
Solution
Z value corresponding to 0.85 probability = 1.0364
Let the mean be M
Then , 1.0364 =( 16-M)/2
Solving for M , we get M = 13.93 ounces
Question
Solution
Z value for 0.9 probability = 1.2816
Hence, 1.2816 = (X -1865)/225
Solving the above, we get X = $2,153.35

Solution
Z = (3000-2210)/460 = 1.717
P (Z>1.717) = 1- P (Z<1.717) = 1- 0.957 = 0.043
Answer is 0.043
Question 4
Requisite probability = 0.5661/(0.1504+0.5661) = 0.7901
Question 5
Event C and R are not independent of each other. This is because the probability of event C is not
the same for the two political affiliations i.e. R and D. If the leader has a R affiliation, then C would
have probability of 0.37 but with D affiliation this probability of C reduces to 0.21. Hence, it is
evident P(C) is a function of the political affiliation of the underlying leader.

Question 6
0.65 is c
0.28 is g
0.20 is a
1 is i
Question
Answer – probability is 0.22
Question 7
Answer is 0.33
Question 8
Answer is 45/365
Question 11

A and B are disjoint
Question 12
Total kittens = 8
Black kittens = 3
Non-black kittens = 8-3 = 5
Thus, in order to ensure that both kittens are non-black, the kittens have to be selected from the 5
kittens.
Hence, requisite probability = (5C2)/(8C2) = (5*4)/(8*7) = 5/14
Question 13
Yes it is possible for A and B to be disjoint since P(A) + P(B) is less than 1. Hence, there is no
guarantee that some portion in the two sets would overlap with certainty.
Question
Solution
Expected number of cases = 1*0.02 + 2*0.03 +3*0.09 +4*0.55 +5*0.2+6*0.06 +7*0.05 = 4.26
Answer is 4.26

Question
Answer is 0.2537
Explanation
Required probability = P(0) + P(1) + P(2) = 25C0 (0.15)0(0.85)25 + 25C1 (0.15)1(0.85)24 +
25C2(0.15)2(0.85)23 = 0.2537
Question

Answer is Option A (No, because 89 is less than one standard deviation)
Question
Answer is 0.2023
Explanation
Z = (16-16.2)/0.24 = -0.8333
P (Z< -0.8333) = 0.2023 (As obtained from z table).
-

Question
Answer is 0.39
Explanation
Requisite probability = P(1) + P(2) + P(3) + P(4) = 0.02 + 0.03 +0.09 + 0.25 = 0.39
Question
Answer

Expected number = 4.26
Question
Mean = np = 55% * 300 = 165
Standard deviation = sqrt (npq) = sqrt (300*0.55 * 0.45) = 8.62