1 Assume that the given procedure yields a binomial distribu

1. Assume that the given procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean and standard deviation . Also, use the range rule of thumb to find the minimum usual value 2 and the maximum usual value +2.

In an analysis of preliminary test results from the a gender-selection method, 42 babies are born and it is assumed that 50% of babies are girls, so n=42 and p=0.5.

The value of the mean is :

2. If a gambler places a bet on the number 7 in roulette, he or she has a 1/38 probability of winning.

a. Find the mean and standard deviation for the number of wins of gamblers who bet on the number 7 two hundred times.

b. Would 0 wins in two hundred bets be an unusually low number of wins?

Solution

1.

As

u = n p = 42*0.5 = 21 [answer, mean]

sigma = sqrt [n p (1 - p)] = sqrt(42*(0.5)*(1-0.5)) = 3.240370349 [answer, standard deviation]

Thus, the usual values lie between

u - 2*sigma < x < u + 2*sigma

21 - 2*3.240370349 < x < 21 + 2*3.240370349

14.5192593 < x < 27.4807407 [ANSWER, USUAL VALUES]

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