normal distribution

As part of a promotion for a new type of cracker, free trial samples are offered to shoppers in a local supermarket. the probability that a shopper will buy a packet of crackers after testing the free sample is 0.20. Different shoppers can be regarded as independent trials. Let X be the number among the next 100 shoppers who buy a packet of the crackers after tasting a free sample, then X has approximately a A. N(20, 16) distribution. B. N(20, 4) distribution. C. N(0.2, 4) distribution. D. N(0.2, 16) distribution. D. N(0.2, 16) distribution. these Q look easy but if i don’t nail them one good last time before the exam … i guess that is the same for u

binomial distribution p=0.2 n=100 sd = 20 var = 100*0.8*0.2=16 ~Z(20,16) A)

oops i mean ev=20, not sd=20

A … this is a good Q. With all focus on Z dist and t-stat, I wasn’t even thinking on the lines of binomial distribution.

Answer should be B, no?

Dreary Wrote: ------------------------------------------------------- > Answer should be B, no? nope, normal is specified by mean & variance

I forgot, is normal distribution N(mean, variance) or N(mean, std)?

Variance for a binomial is np(1-p) so 16 it is.

sure that’s A