A fair coin is flipped 100 times. Use a normal approximation to find the probability of 60 or more heads.
1
Find the expected count
X counts heads in n=100 independent flips with p=0.5, so X∼B(100,0.5). The normal curve to use has the binomial’s own mean and standard deviation.
E(X)=100⋅0.5=50
2
Find the variance
Multiply by the chance of a tail, 1−p=0.5, as well.
Var(X)=100⋅0.5⋅0.5=25
3
Take the square root
The standard deviation is in heads, like the count.
√25=5
4
Check that a normal curve is reasonable
np=50 and n(1−p)=50. Both are well above 10, so the distribution is close to a bell shape with mean 50 and standard deviation 5.
5
Apply the continuity correction
On the continuous scale the count 60 occupies the interval from 59.5 to 60.5, so “60 or more” starts at 59.5. Standardize that point.
z=559.5−50=1.9
6
Find the tail area
The area to the right of z=1.9 under the standard normal curve is 1−Φ(1.9), and a table gives Φ(1.9)≈0.9713.
1−0.9713=0.0287
7
Compare with the exact value
Adding the binomial probabilities from k=60 to 100 gives 0.0284, so the approximation is off by 0.0003. Without the correction, z=(60−50)/5=2 and the area is 0.0228, noticeably too small.
Result
About 0.0287, or 2.9%. The exact binomial probability is 0.0284.
Your turn
1. One like the example
For the same 100 flips, approximate the probability of 45 or fewer heads.
Show the answer and explanation
About 0.184.
With the continuity correction, “45 or fewer” ends at 45.5, so z=(45.5−50)/5=−0.9 and Φ(−0.9)≈0.1841. The exact binomial probability is 0.1841 as well. The match is this close because p=0.5 makes the distribution symmetric.
In a large city, 30% of voters support a measure. In a random sample of 200 voters, approximate the probability that at least 70 support it.
Show the answer and explanation
About 0.071.
np=60 and n(1−p)=140, both well above 10. The standard deviation is √42≈6.48. “At least 70” starts at 69.5, so z≈1.47 and the table gives 1−0.9292=0.0708. Software with the unrounded z gives 0.0713, and the exact binomial sum is 0.0728.
A factory’s chips are defective with probability 0.02, independently. A box holds 100 chips. Should a normal curve be used for the chance of no defects in a box? Find that chance.
Show the answer and explanation
No, because np=2 is far below 10. The exact chance is 0.98100≈0.1326.
With an expected count of 2, the distribution is piled up near 0 and skewed right, not bell-shaped. No defects means 100 good chips in a row, (0.98)100, which the binomial formula gives directly.
X∼B(100,0.02)100⋅0.02=20.98100≈0.1326
Keep exploring
In Statistics, Distributions shows P(X≥1.9)=0.0287166 for the standard normal. Change the value to 2, the z-score without the continuity correction, and the tail falls to about 0.0228.
