Let Sn be the number of successes in n independent Bernoulli trials, where the probability of success at each trial is 1/3. Provide a numerical value, to a precision of 3 decimal places, for each of the following limits. You may want to refer to the standard normal table.
1. lim n→∞ P(n/3−10≤Sn≤n/3+10)=
2. lim n→∞ P(n/3−n/6≤Sn≤n/3+n/6)=
ans = 1
3. lim as n→∞ P(n/3−√2n/5≤Sn≤n/3+√2n/5)=
Kindly help with question 1 & 3.
1) 0
2) 1
3) 0.4514
To find the numerical values for limits 1 and 3, we can use the Central Limit Theorem, which states that as the number of trials (n) increases, the distribution of Sn approaches a normal distribution with mean np and standard deviation √(np(1-p)), where p is the probability of success.
1. For limit 1:
We have n approaching infinity and the probability of success p = 1/3.
Using the Central Limit Theorem, we can approximate the probability as:
P(n/3 - 10 ≤ Sn ≤ n/3 + 10) = P((n/3 - np)/√(np(1-p)) ≤ (Sn - np)/√(np(1-p)) ≤ (n/3 + 10 - np)/√(np(1-p)))
Since n is approaching infinity, we have:
[√(n/3) - √(np(1-p))]/√(np(1-p)) ≤ (Sn - np)/√(np(1-p)) ≤ [√(n/3) + √(np(1-p))]/√(np(1-p))
We can now compute the values for the limits:
lim n→∞ [√(n/3) - √(np(1-p))]/√(np(1-p)) = -√(2/3) ≈ -0.816
lim n→∞ [√(n/3) + √(np(1-p))]/√(np(1-p)) = √(2/3) ≈ 0.816
Therefore, lim n→∞ P(n/3 - 10 ≤ Sn ≤ n/3 + 10) ≈ P(-0.816 ≤ Z ≤ 0.816), where Z is a standard normal random variable.
Using the standard normal table, we can find the probability associated with Z = 0.816:
P(-0.816 ≤ Z ≤ 0.816) ≈ 0.792
Hence, lim n→∞ P(n/3 - 10 ≤ Sn ≤ n/3 + 10) ≈ 0.792.
3. For limit 3:
Using a similar approach as in limit 1, we have limit as n approaches infinity for the probability:
P(n/3 - √(2n/5) ≤ Sn ≤ n/3 + √(2n/5))
Apply the Central Limit Theorem to approximate the probability:
lim n→∞ P((n/3 - np)/√(np(1-p)) ≤ (Sn - np)/√(np(1-p)) ≤ (n/3 + √(2n/5) - np)/√(np(1-p)))
Simplifying the expression gives:
lim n→∞ [√(n/3) - √(np(1-p))]/√(np(1-p)) ≤ (Sn - np)/√(np(1-p)) ≤ [√(n/3) + √(2n/5) + √(np(1-p))]/√(np(1-p))
Evaluate the limits:
lim n→∞ [√(n/3) - √(np(1-p))]/√(np(1-p)) = -√(2/3) ≈ -0.816
lim n→∞ [√(n/3) + √(2n/5) + √(np(1-p))]/√(np(1-p)) = √(2/3) ≈ 0.816
Therefore, lim n→∞ P(n/3 - √(2n/5) ≤ Sn ≤ n/3 + √(2n/5)) ≈ P(-0.816 ≤ Z ≤ 0.816), where Z is a standard normal random variable.
Using the standard normal table, we can find the probability associated with Z = 0.816:
P(-0.816 ≤ Z ≤ 0.816) ≈ 0.792
Hence, lim n→∞ P(n/3 - √(2n/5) ≤ Sn ≤ n/3 + √(2n/5)) ≈ 0.792.
To calculate the limits in questions 1 and 3, we can use the properties of the normal distribution and the central limit theorem.
1. To find the limit in question 1, we need to calculate the probability of Sn falling within the range [n/3 - 10, n/3 + 10] as n approaches infinity.
The probability distribution of Sn follows a binomial distribution with parameters n and p, where p is the probability of success in each trial. In this case, p = 1/3.
As n approaches infinity, the binomial distribution approaches a normal distribution. The mean of the binomial distribution is np, and the standard deviation is sqrt(np(1-p)).
Therefore, the mean of Sn is n/3 and the standard deviation is sqrt(n/3 * (1 - 1/3)).
To calculate the required probability, we need to convert the range [n/3 - 10, n/3 + 10] into standard units.
The lower limit in standard units is (n/3 - 10 - n/3) / sqrt(n/3 * (1 - 1/3)) = -10 / sqrt(n/3 * (2/3)).
Similarly, the upper limit in standard units is (n/3 + 10 - n/3) / sqrt(n/3 * (1 - 1/3)) = 10 / sqrt(n/3 * (2/3)).
From the standard normal table, you can find the probability of a standard normal distribution falling between these z-scores.
2. To find the limit in question 3, the approach is similar. We need to calculate the probability of Sn falling within the range [n/3 - √(2n/5), n/3 + √(2n/5)] as n approaches infinity.
Again, we convert the range into standard units using the mean and standard deviation of Sn.
The lower limit in standard units is (n/3 - √(2n/5) - n/3) / sqrt(n/3 * (1 - 1/3)) = -√(2n/5) / sqrt(n/3 * (2/3)).
Similarly, the upper limit in standard units is (√(2n/5) - n/3) / sqrt(n/3 * (1 - 1/3)) = √(2n/5) / sqrt(n/3 * (2/3)).
Again, you can use the standard normal table to find the desired probability.
Please note that these calculations involve approximations and rely on assumptions made by the central limit theorem.