the normal approximation to the binomial distribution

Question

a magazine reported that 6% of american drivers admit to reading a newspaper while driving. if 500 drivers are selected at random, find the probability that exactly 40 will admit to reading the newspaper while driving.

Answer 1

Obtain z values to perform the normal approximation. Take the average of the sample as μ equal to np or (500)(0.06) or 30. Take the standard deviation of the sample as √(npq) or √(500×0.06×0.94) or √28.2.

Next bracket 40 between 39.5 and 40.5 by computing (39.5 −30)/√28.2 or 1.788953496 and (40.5 −30)/√28.2 or 1.97726439.

A highly accurate calculator program gives respective proportions of area under the standard normal curve for

1.788953496 & 1.97726439 as 0.4631888474 and 0.4759941237. Then (0.4759941237 − 0.4631888474) or 0.01280527633 gives the normal approximation sought.

Now find the actual probability by computing [500! ÷ 40! ÷ (500 − 40)!] times 0.06⁴⁰ times 0.94^{(500 - 40)} . A calculator with large factorial capacity reduces [500! ÷ 40! ÷ (500 − 40)!] times 0.06⁴⁰ times 0.94^{(500 - 40)} to

0.01305983570921287575347173747943.

Divide 0.01280527633 by 0.01305983570921287575347173747943 to yield 0.98050822499755487336187665395358 which indicates that the approximate solution is 98.05% accurate.

Note that the approximation will be acceptably close when both np and nq are greater than or equal to 5.