Question

1. (a) Consider the random variable Y having possible values 1, 2 and 3. The corresponding probability for each value is: 1 with probability Y = 2 with probability 3 with probability Determine an expression for the probability mass function (pmf) (11) Determine the mean and the standard deviation of Y. (b) The probability that a man hits a target is and that of his son and daughter are and respectively. If they all fire together, find the probability that: (1) they all miss the target exactly one shot the target at least one shot the target (c) An insurance company claims that full size cars are more prone to automobiles accidents and hence should be subject to a higher insurance premium. To test the validity of the claim, an auto firm gathered a random sample: Car Size No Accident At Least One Accident 24 34 13 117 214 108 Full Size Compact Small Formulate the necessary hypotheses and test at 1% level of significance that cars sizes and occurrence of accidents are independent. a) A random variable k has the probability function, b P(k) = k = 1,2,3.. 2kk Compute the value of b.

Answer 1

T 6 1 т (a)(1) P(X = r) 2 = 1,2,3 (2) E(x²) = < <?P(X = x ) = 1?(1/6) + 2? (2/6) +3²(3/6) = 6 E(X)= XP(X = x ) = 1(1/6) + 2(2/6) + 3(3/6) = 2.33 V(X) = E(X) – (E(X))2 = 0.571 (6) P(D) = 1/2; P(S) = 2/5; P(d) = 1/5 (1) P(all miss the target) = P(Dºn Sºnd) = P(DC)P(SC)P(dº) ;since all the events are independent of each other. = (1 – 1/2)(1 – 2/5)(1 – 1/5) = (1/2)(3/5)(4/5) = 6/25 (2) P(exactly one shot) = P((DNS nd)U(Dºnsnd)U(Dº n se nd)) = P((Dn Sºnd) + P(Disnd) + P(Dºn Sºnd) since these are mutually exclusive cases. = P(D)P(S) P(d) + P(D)P(S) P(d) + P(D)P(S) P(d) = (1/2)(1-2/5)(1-1/5)+(1-1/2)(2/5)(1-1/5)+(1-1/2)(1-2/5)(1/5) = 23/50 (3) P(alteast one shot the target) = 1 - P(none hit the target) = 1 – 6/25 19/25 (d) there are total 37 full size car and we want to test whether the probability of accident by full size car is more than than others. Ho: P1= P2
$H_1:P_1-P_2>0\\ Z=\frac{p_1-p_2}{\sqrt{p_1q_1/n_1+p_2q_2/n_2}}\\ $value of test statistic is$\\ Z=\frac{24/37-34/151}{\sqrt{0.0061596+0.0011554}}=4.95\\ $we reject the null hypothesis if $Z>2.58$otherwise not.$\\ $hence calculated value is greater than 2.58 hence we reject the null hypothesis.$\\ H_1:P_1-P_3>0\\Z=\frac{24/37-108/322}{\sqrt{0.0068518}}=3.78\\ $we reject the null hypothesis if $Z>2.58$otherwise not.$\\ $hence calculated value is greater than 2.58 hence we reject the null hypothesis.$\\ $Hence probability of accident for fullsize car is more than others.$$
k k (5) ΣΡ(K = k) = 1 Σό/2* = 1 = 6(1/2 + 1/22 + 1/23 +....) = 1 1 και (1/2) (G 1 - 1/2 + b = 1 since 1 1+1/2 + 1/22 + 1 – 1/2 1