During the 2009 edition of the reality show Britain’s Got Talent, runner-up and Internet singing sensation Susan Boyle obtained 20.2% of the first-place votes. Suppose that this percentage would hold...

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Part A
P (X = 57), n = 250, p = 0.202
X ∼ Binomial(n = 250, p = 0.202)
P (X = 57) =
(
n
x
)

px(1−p)n−x =
(250
57
)
(0.202)57(1−0.202)250−57 = 1.1642897196156E+
57 ∗ (0.202)57(0.798)193 = 0.036112 ≈ 0.036112 ≈ 0.036
P (X = 57) = 0.036
Using excel function BinomDist(57,250,0.202,false) or TI-83/84 function
binompdf(250,0.202,57), exact answer is 0.036111635
We can also solve this using normal approximation to binomial random variable.
Binomial can be approximated to normal with:
µ = np = 250 ∗ 0.202 = 50.5
σ =
√
np(1 − p) =
√
250 ∗ (0.202)(1 − 0.202) = 6.348149 ≈ 6.348
P (X = 57) Using normal approximation
Since we are approximating a discrete distribution by continuous normal distri-
bution, values between 56.5 and 57.5 both approximate to 57. Thus, “equal to 57
” corresponds to continuous normal distribution with P(56.5 < X < 57.5) after
continuity correction.
Normal Distribution, x1 = 56.5, x2 = 57.5, µ = 50.5, σ = 6.348
We convert this to standard normal using z = x− µ
σ
z1 =
56.5 − 50.5
6.348 ≈ 0.945157 ≈ 0.95
z2 =
57.5 − 50.5
6.348 ≈ 1.102684 ≈ 1.10
P (56.5 < X < 57.5) = Area in between 56.5 and 57.5
X
µ = 50.5
σ = 6.348149P (X < 56.5)
57.5
56.5
50.5
P (X < 57.5)
P (56.5 < X < 57.5)
P (56.5 < X < 57.5) = P (0.95 < Z < 1.10)
= P (Z < 1.10) − P (Z < 0.95)
= 0.8643 − 0.8289 (from z-table)
= 0.0354
P (X = 57) ≈...