Question

Please answer Q3 and Q4.

Question 3 An insurance company offers a life insurance of $200 000 valid for one year. Its cost is $250. This insurance is available only for healthy people between 18 and 35 years old. According to the actuarial tables, the probability that someone of the given characteristics dies is of 0.05%. (a) Determine the expected value as well as the standard deviation for the gain obtained by selling one life insurance of this type. Interpret your results. (b) If the insurance company sells 2 500 life insurances of this type, give the probability the company makes a deficit Question 4 The manager of a restaurant analyzes the duration of his client's visits. He estimates that a client stays at the restaurant for 85 minutes on average, and with standard deviation of 25 minutes. Assume that the duration of a visit follows a normal distribution (a) Give the probability that a client stays at the restaurant for less than an hour (b) Give the probability that a client stays at the restaurant between 70 and 100 minutes. (c) Give the probability that 10 individual clients stay at the restaurant between 70 and 100 minutes, in average (d) The manager wants to know the 1st quartile as well as the 3rd quartile of the duration of his clients visits. What are these values?

Answer 1

Here we have given that a client stays at the resturent avenage time 85 mins and Standard deviation 25 minutes. as it follows normal distribution. o Lt, & be the varriaste which clinant stays the reastaurent. defines that the a ie, XN N(85, 25) stays at the restaurant fers Probability less than " that a client one houn is Comins) 2-85 than one hour is P(x460) = P(x- 85 & 60-85 -PTZ 4-1 = 9(-1) 60-85 25 0.8413 - a form Where z is the Standard Normal Vorriate and 20) is 1- P1 - 1-0.8413 1 2 (1) = 0.1587 nomal table the caf of Z (6) Probability that client Stays between 20 e 100 mins = 9(06) - 960-c) = P(0.6) - (-9 (0-16) = 2 | (0.6) - 1 = 2*0.726-1 = 1.452-1 = 0.452

& If §i ad fo be by defination the first and 300 quartile then P(x481) = 0.25 aw P(x *93) = 0.25 0.25 x=0, n=9 When x = b j = 93-4 = 2 (say) (1) 8:25 10.80.25 0.25 x = 81.; Z = 91-4 - z say) - (2) I x= u bo substraiting (1)-(B) we get, 22-93-31 2= -2, Z=0 Z=2 Now from the figure obviously PloKZ052) = 0.25 => 2,20.67 from normal table So from h 93-M 0.67 (first quartile) 8 = 0.67.8tu = 0.678 25+ 85 = 101.75 mins and from (2) Bill = -21 (3rd quantile) > 813 M-r2i = 85- 25x0,67 = 68.25 mins from normal