If X ~ B(n, p) and if n is large and/or p is close to ½, then X is approximately N(np, npq)
Here n is large so normal distribution can be used as an approximation to the binomial distribution.
Step 1: find mean and sd
mean = np= 400*0.25=100
sd= s="\\sqrt{np(1-p)}" ="\\sqrt{400(0.25)(1-0.25)}" = 8.66
Step 2: since np=400*0.25= 100
nq=400*(1-0.25)= 300, both are greater than 10, we use continuity correction factor.
Step 3:
Use the continuity correction factor on the X value.
if P(X>a) use P(x>a+0.5)
Hence P(X>110)=P(X>110.5)[using continuity correction factor)
Step4: find z score
Z=(x-mean)/sd
Z= (110.5-100)/8.66
Z=1.1124
Step 5:
so P(Z>1.2124)=1-0.8873= 0.1127
so then, we conclude that P( X >110 ) approx 0.1127
P(X>110)≈0.1127, which ends the calculation of the requested probability
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