TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Levajkovic)/Übungen 2023W/HW05.4

Coin throws

An unfair coin is thrown 600 times. The probability of geting a tail in each throw is ${\frac {1}{4}}$.

(a) Use a Binomial distribution to compute the probability that the number of heads obtained does not differ more than 10 from 440.

(b) Use a Normal approximation without a continuity correction to calculate the probability in (a). How does the result change if the approimation is provided with a continuity correction?

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Ähnliches Beispiel: Übungen 2020W - HW 5.2

Lösungsvorschlag von Simplex

# a)
throws <- 600
prob <- 3/4

res2a <- pbinom(450, throws, prob) - pbinom(430, throws, prob)
print(res2a)

# b)
exp <- 0
sd <- 1
a <- 430
b <- 450

res2b_without_correction <- 
    pnorm((b - throws * prob)/(sqrt(throws * prob * (1-prob)))) - 
    pnorm((a - throws * prob)/(sqrt(throws * prob * (1-prob))))
res2b_with_correction <- 
    pnorm((b + 1/2 - throws * prob)/(sqrt(throws * prob * (1-prob)))) - 
    pnorm((a - 1/2 - throws * prob)/(sqrt(throws * prob * (1-prob))))

print(res2b_without_correction)
print(res2b_with_correction)

(a)

$$X\sim B(600,1/4):P(430<X\leq 450)=P(X\leq 450)-P(X\leq 430)=\sum _{x=430}^{450}{\binom {600}{x}}(1/4)^{x}\cdot (3/4)^{600-x}\approx 48.14\%$$

(b)

Calculate with correction: fill in sigma $\pm 1/2$

$${\text{Standardization: }}LetX\sim {\mathcal {N}}(\mu ,\sigma ^{2}):Z={\frac {X-\mu }{\sigma }}\sim {\mathcal {N}}((0,1)$$

• Without correction: 47.03%

• With correction: 49.22%

--Simplex 14:04, 3. Feb. 2023 (CET)

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