Question

4. The company Crunchy Hippy Granola Inc. anticipates that the num- ber of new customers (Y) resulting from an intensive marketing cam- paign in the Chicago area will follow (approximately) a log-normal distribution with mean ?? 800 and variance ?2-512. It is known that X-lny follows a normal distribution where ? and ?2 denote the mean and variance of (a) Why should the lognormal distribution only be considered an ap- proximation here? (Consider the variable being measured and comment. (b) Find the mean and varia (c) Calculate P(Y ? 600), the probability that Hipster Granola, Inc (d) Find the 40th and 95th percentiles of Y nce of the normal random variable (Hint: it may be convenient here to work with CVY-?y/? will have at least 600 new customers in the Chicago area .

Answer 1

The log.normal distribution is the distribution that can flexibly takes any non-negative values. Alse, log-normal distribution can take positive values of both integers and fractions. According to the question, Y is the number of new customers resulting from campaign, so Y is a positive integer here. Therefore, log-nomal is considered an approximation in this case Consider Y be the number of customers following Lognormal distribution with, Mean(4) 800 variance(q)-5. The mean of Y can be computed as follows: 800 expl μ+ The variance of Y can.be.computed as follows:

Now, find the mean of X as follows: σ2 expl μ+- 800 Squaring both sides,

expl μ+- (800) exp | 21 μ+- (800) exp(2μ + σ*) ny -ap(o)-1 exp (800) 12 exp (σ)-3824697 ơ-in (382.4697) σ = 5.95 Now, compute the variance by substituting the value σ. 5.95 in the formula of mean of Log normal distribution as follows

5.95 800 = expl μ +-- In (800) = μ + 2.97 μ=668-297 = 3.71 Conclusion: The mean and variance of normal random variable Xis 3.71 and 5.95 respectively

Compute the probability that Hipster Granola Inc. will have at least 600 new customers in Chicago area as follows: P(Y>600) 1-P(Y <600) Since, Y follows Log-normal distribution, therefore, P(>600) 1-P(In() <In (600) 1-P(In(r)-μ ln (6002 In (600)-3.71 V5.95 = 1-P(z <1.10) Using standard normal table, the value of z-10 is 0.8643 P(> 600) 1-0.8643 0.1357

Conclusion: There are approximately 14% chances that Hipster Granola Inc. will have at least 600 new customers in Chicago area. Compute the 40th percentile of I as follows: P(Y <y) 0.4 (v)-0.4 = 0.4 P In(Y) In Using standard normal table, find the value of z with the probability nearest to 0.4, that is, 0.25. Therefore, P(z <-0.25) 0.4 From equation (1) and (2), it çan beconcluded that

-0.25 In(y)-3.71_-025 V5.95 5.95 In (y)-0.25x V5.95) +3.71 exp (3.1002) у 22.20 У~22 Conclusion: The 40h percentile of new customers is approximately 22 Compute the 95th percentile of Y as follows PO' < y) = 0.95 P[In (X) <ln()0.95 (In (T)-< In(y) -0.95 In (y)-| = 0.95

Using standard normal table, find the value of z with the probability nearest to 0.9, that is, 1.645. Therefore, P(z <1.645) 0.95 From equation (1) and (2), it çan beconcluded that In()-H-1645 1.645 In()-3-71-1.645 = 1.645 V5.95 In (y)-(1.645xV5.95) +3.71 y exp (7.7226) , 2258.82 У ~ 2259

Conclusion: The 956 percentile of new customers is approximately 2259