Question

You are playing a card came, and the probability that you will win a game is p = 0.39. If you play the game 95 times, what is the most likely number of wins? (Round answer to one decimal place.) = Let X represent the number of games (out of 95) that you win. Find the standard deviation for the probability distribution of X. (Round answer to two decimal places.) The range rule of thumb specifies that the minimum usual value for a random variable is J-20 and the maximum usual value is u+20. You already found us and o for the random variable X. Use the range rule of thumb to find the usual range of X values. Enter answer as an interval using square-brackets and only whole numbers. usual values

Answer 1

Solutions Where x is no.op wins, hue xn B[n, p=0.39). 4 Xn B( m, p). and then, Hh or F(x)= mp and var(x)= npq, where q, is failue. cott 9 = 1-21 here, in this case total no.O game plavjed is 95 times So, n=95 then mean ore= np = 95% 0.39 = 37.05 - 37.1 * Find the standard deviation for the probability distibution qoX ? Variance of X or 02 = npq = nxpx (1-P) = 95%.39 8.61 = 22.6 Standard dencatten or 6 = Vvariance = j22.6 : = 4.691 = 4.69

Now muchane mian um 37.05 and slendard deviation 6 4.691 "Vange of X is equals [jk-at, ultar] - [37.05=Rx4.682), 37.05+ (2x4.67) = [ 34.05-(9.362), 31.05+ (9.382)] = [27.67, 4643] in [28, 46] Solution: where x is the no.of wins or success. Xu B(n,p=.26) If au BC n,p) mean pl = np and var (x) = npq, where 9-1-P. ns rio.of trails . n=1468. mean =) M= np . = 1468 x .26 = 381.68 = 381.7 !! Standard deviatin = V variance (x). = Inpa = 11468X226 X.74 = T282.4432 = 16.81 o

range of x= [u-as , 4+25 mee our mean u= 381.7., and standard dewaitin on = 16.81 = 361.4- (ax 1681), 3815 +(a X 16:87] = [381.4–(33.62), 361.7 +(33.60) = [348.08, 41534] 6 [348, 415)