Question

Question 7 1 pts For a binomial distribution with n = 396 and p = 0.45, determine the mean and standard deviation you would use when approximating probabilities with a normal distribution. O = 217.8 and o = 14.758 Ou = 178.2 and = 13.349 OH = 98.01 and 0 = 9.9 O = 217.8 and 0 = 98.01 Op=178.2 and 0 = 99 u=178.2 and 0 = 98.01

Answer 1

[Q.77 Binomial Distribution n=396 P = 0.45 .q=1-P = 1-0.45 = 0.55 Aproximating Binomial distribution to Normal, u= np & 0²= npq :u=390 (0.45) 62396 (0.45) (o.ss) u=178.2 621= 98.00 10 J98.01 u = 178.2 6 = 9.899 6= 9.9 6 ~ 9.9. 5th option is correct? [Q.8 / let X be the no. of all new products in Grocesy stores fall within 2 years.& p be the proportion of them. P = 0.8 n = 36. P ( X = 30) = To find. . X . Bin (3608) :-P (X=30) = nexpo (1-P) 36-30 - 36 c, 10.8150 (1-0.8) =(59024) (0.00124) (0.0000647 = 011543 Aproximately option 6th is correet. ret X be the no.ot defective baloons in a shipment. n=300. . .. PC Accepting shipment) P=0.05 = P ( X 510) = P(X=o) + P(x=1)... + P(X=100 Aproximating Binomial to normal distribution. P ( x < 10) = P (X-P & 10-300 (0.05). I Jnpa V 300 (Oos (0.95) - PC zs -1.327 = 0,0834 option # ist is correct.

Let x be the no. of new products Among 4g new one, those fall within 2 days. . X . Binomial disth n = 49 P = 08 To find P(X=40) - P(x = 40) + P(X=41)+ ... + P(X=49) I using Binomial probability table P(X=40 ) = 0.4717 Aproximately option last one is corret.