Normal Approximation of Binomial Distribution When n 30 and...

Normal Approximation of Binomial Distribution When n 30 and np 5; then mean = std. dev. = Question 1: For a large group of sales prospects, it has been observed that 20% of those contacted personally by a sales representative will make a purchase. If a sales representative contacts 40 prospects, use a suitable approximation to determine the probability that (a) 10 or more will make purchase (b) More than 11 will make purchase (c) At most 10 will make purchase (d) Fewer than 11 will make purchase.

Transcript

00:01 Hello student given that a binomial distribution has base equals to 0 .20 and given that n is equals to 100 then we have to find out the mean and standard deviation.

00:15 So we know that mean of binomial distribution is equals to n into p.

00:22 So here 100 into 0 .20 therefore mean is equals to 20.

00:29 Now we have to find out where standard deviation.

00:34 So here standard deviation is the square root of variance.

00:38 So we know that variance of binomial distribution is npq.

00:41 So here square root of npq which is equal to square root of 16 therefore standard deviation is equals to 4.

00:50 Now binomial probabilities can be approximated by normal.

00:55 So here n greater than or equal to 30.

01:00 Now we check for second condition that is np greater than or equal to 5 or not.

01:07 So here np is equals to 20 which is greater than or equal to 5 and now we check for n into 1 minus p that is nq greater than or equal to 5.

01:21 So here nq also greater than or equal to 5.

01:24 So here answer is yes because n greater than or equal to 30 and p greater than or equal to np greater than or equal to 5.

01:37 Now here we have to find out the probabilities of exactly 23 successes.

01:43 So here we use normal approximation to binomial.

01:48 So here mu is equals to np and sigma is equals to square root of npq therefore sigma is equal to square root of 4...

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