Suppose a fair coin is tossed n times. The probability of obtaining head and tail are the same because this is a fair coin. The proportion of heads is defined as the number of heads appeared divided by n. We can model this probability distribution by binominal distribution.

For large n, the binominal distribution can be replaced by the normal distribution with μ=np and δ=√np(1-p)
Suppose we would like to obtain the proportion of heads between 0.45 and 0.55 for a fair coin tossed n times. That means we are trying to find: P(0.45n≤x≤0.55n)

a. are the conditions satisfied for replacing binominal probability distribution by normal distribution if n=100?
b. Find the numerical value of P(0.45n≤x≤0.55n) for n=100 by normal distribution if conditions in (a) have been satisfied.
c. Find the general formula of cumulative distribution function φ in standard normal distribution as a function of n.
d. Show that when n increases, the continuity correction becomes less and less important
e. Verify the claim in (d) by using n=100, 250, 500, 750 and 1000

Solution Summary

The solution deals with the continuity correction of binomial distribution to Normal distribution.

... (d) more than 9 but fewer than 19 employees steal? Please see the attachment. Using the continuity correction and standardizing, this solution is solved. ...

... Considering the continuity correction, and Z score P(100≤X≤110)=P(99.5<X<110.5) =P(-2.959<Z<-1.3712)=0.0836 Probability for a Range From X Value 99.5 To X ...

... Answer: Mean= 52.2 Standard deviation= 5.00. b) One half of the sample = 1/ 2 x 100 = 50 We will apply continuity correction factor 50-1/2= 49.5. ...

... each answer, then the probability that she gets any one answer correct is 0.2.) Use the normal approximation to the binomial with a correction for continuity. ...

... npq = 49.02. √npq = 7.001. since the continuity correction factor is to be used we have. P( Xb ≤ x) ≈ P( Xn ≤ (x + 0.5) ). P(x≤100) = P( z ≤100.5). ...