The random variable X ≈ B(n, p)
Where n is the number of samples. If the value of n is huge, calculating binomial coefficient can be difficult and calculator won't be able to perform the nCr calculation for a big value of n (try ⁶⁰⁰C₁₂₃ on the calculator).
If p, which is the probability, is close to 0.5, the binomial distribution will be fairly symmetric.
We have n = 114 and p = 22% = 0.22
We first need to approximate X by Y ≈ N(np, np(p-1))
np = 114×0.22 = 25.08
np(1-p) = 114×0.22×0.78 = 19.6
Therefore, Y ≈ N(25.08, 19.6)
and since the standard deviation, б = √np(1-p) = √19.6
We then have, Y≈N(25.08, (√19.6)²)
We aim to find the probability of X is exactly 27, mathematically written as P(X=27)
Applying the continuity correction we have P(26.5<Y<27.5) ⇒ This correction, in other words, is applying a range to X that will definitely include the value of X=27
The next step is standardising
We have the mean, μ=25.08 and σ=√19.6
P(26.5<Y<27.5) = P(
<Z<
)
P(26.5<Y<27.5) = P(0.32<Z<0.55)
P(26.5<Y<27.5) = P(Z<0.55) - [1 - P(Z<0.32)] ⇒ Read the values for these z-score on the z-table
P(26.5<Y<27.5) = 0.7088 - [1-0.6255]
P(26.5<Y<27.5) = 0.3343
Where n is the number of samples. If the value of n is huge, calculating binomial coefficient can be difficult and calculator won't be able to perform the nCr calculation for a big value of n (try ⁶⁰⁰C₁₂₃ on the calculator).
If p, which is the probability, is close to 0.5, the binomial distribution will be fairly symmetric.
We have n = 114 and p = 22% = 0.22
We first need to approximate X by Y ≈ N(np, np(p-1))
np = 114×0.22 = 25.08
np(1-p) = 114×0.22×0.78 = 19.6
Therefore, Y ≈ N(25.08, 19.6)
and since the standard deviation, б = √np(1-p) = √19.6
We then have, Y≈N(25.08, (√19.6)²)
We aim to find the probability of X is exactly 27, mathematically written as P(X=27)
Applying the continuity correction we have P(26.5<Y<27.5) ⇒ This correction, in other words, is applying a range to X that will definitely include the value of X=27
The next step is standardising
We have the mean, μ=25.08 and σ=√19.6
P(26.5<Y<27.5) = P(
P(26.5<Y<27.5) = P(0.32<Z<0.55)
P(26.5<Y<27.5) = P(Z<0.55) - [1 - P(Z<0.32)] ⇒ Read the values for these z-score on the z-table
P(26.5<Y<27.5) = 0.7088 - [1-0.6255]
P(26.5<Y<27.5) = 0.3343