Let Sn be the number of successes in n independent Bernoulli trials, where the probability of success for each trial is 1/3 . Provide a numerical value, to a precision of 3 decimal places, for each of the following limits.
1. lim P (n/3 - 10 < Sn < n/3 + 10)
2. lim P (n/3 - n/6 < Sn < n/3 + n/6)
3. lim P (n/3 - sqrt(2n)/5 < Sn < n/3 + sqrt(2n)/5)
To solve these problems, we can use the Central Limit Theorem, which states that the distribution of the sum (or average) of independent and identically distributed random variables will approach a normal distribution as the sample size increases.
In this case, Sn represents the sum of n independent Bernoulli trials, where each trial has a probability of success (p) equal to 1/3.
1. To find the limit:
lim P(n/3 - 10 < Sn < n/3 + 10)
We can rewrite this limit in terms of the standard deviation using the formula for the standard deviation of a binomial distribution, which is sqrt(n * p * (1 - p)).
The standard deviation (σ) for each Bernoulli trial is: σ = sqrt(n * (1/3) * (1 - 1/3)) = sqrt(2n)/3.
Now, we need to find the number of standard deviations that represent the difference between n/3 - 10 and n/3 + 10. The difference is 20, which is equivalent to 20 / (sqrt(2n)/3) standard deviations.
The limit can be calculated using a z-table or a calculator that gives probabilities for a standard normal distribution. By finding the area between -20 / (sqrt(2n)/3) and 20 / (sqrt(2n)/3), we can find the answer.
2. To find the limit:
lim P(n/3 - n/6 < Sn < n/3 + n/6)
Similar to the previous question, we need to find the number of standard deviations that represent the difference between n/3 - n/6 and n/3 + n/6. The difference is n/6, which is equivalent to n/6 / (sqrt(2n)/3) standard deviations.
Using the z-table or a calculator, finding the area between -n/6 / (sqrt(2n)/3) and n/6 / (sqrt(2n)/3) will give us the answer.
3. To find the limit:
lim P(n/3 - sqrt(2n)/5 < Sn < n/3 + sqrt(2n)/5)
Again, we need to find the number of standard deviations that represent the difference between n/3 - sqrt(2n)/5 and n/3 + sqrt(2n)/5. The difference is 2sqrt(2n)/5, which is equivalent to 2sqrt(2n)/5 / (sqrt(2n)/3) standard deviations.
By calculating the area between -2sqrt(2n)/5 / (sqrt(2n)/3) and 2sqrt(2n)/5 / (sqrt(2n)/3) using the z-table or a calculator, we can find the answer.
Remember to substitute the appropriate values of n into the formulas before calculating the limits.
To find the limits in each of these expressions, we will use the Central Limit Theorem (CLT), which states that the sum of a large number of independent and identically distributed (i.i.d.) random variables with finite mean and variance approaches a normal distribution.
Let's calculate the limits:
1. lim P(n/3 - 10 < Sn < n/3 + 10)
We can rewrite the expression as:
lim P(-10 < Sn - n/3 < 10)
Since Sn follows a binomial distribution with n trials and a probability of success of 1/3, the mean of Sn is n * (1/3) = n/3, and the variance is n * (1/3) * (2/3) = 2*n/9.
According to the CLT, we can approximate Sn with a normal distribution with mean n/3 and variance 2*n/9.
Therefore, the expression becomes:
lim P(-10 < (Sn - n/3) / √(2*n/9) < 10)
This is equivalent to:
lim P(-10 * √(2*n/9) < Sn - n/3 < 10 * √(2*n/9))
Given that n approaches infinity, the standard deviation (√(2*n/9)) will also approach infinity.
So, as n approaches infinity, the probability approaches 1, because the width of the range (-10 * √(2*n/9), 10 * √(2*n/9)) becomes arbitrarily large.
Therefore, the limit is 1.
2. lim P(n/3 - n/6 < Sn < n/3 + n/6)
Similar to the previous calculation, we can rewrite the expression as:
lim P(-n/6 < Sn - n/3 < n/6)
Using the same reasoning and approximating Sn with a normal distribution with mean n/3 and variance 2*n/9, the expression becomes:
lim P(-n/6 * √(2*n/9) < Sn - n/3 < n/6 * √(2*n/9))
Again, as n approaches infinity, the standard deviation (√(2*n/9)) will also approach infinity.
So, as n approaches infinity, the probability approaches 1, because the width of the range (-n/6 * √(2*n/9), n/6 * √(2*n/9)) becomes arbitrarily large.
Therefore, the limit is 1.
3. lim P(n/3 - sqrt(2n)/5 < Sn < n/3 + sqrt(2n)/5)
Using the same approach, the expression becomes:
lim P(-sqrt(2n)/5 < Sn - n/3 < sqrt(2n)/5)
Approximating Sn with a normal distribution with mean n/3 and variance 2*n/9, the expression becomes:
lim P(-sqrt(2n)/5 * √(2n/9) < Sn - n/3 < sqrt(2n)/5 * √(2n/9))
As n approaches infinity, the standard deviation (√(2n/9 * 2n/9)) will approach infinity as well.
So, as n approaches infinity, the probability approaches 1, because the width of the range (-sqrt(2n)/5 * √(2n/9), sqrt(2n)/5 * √(2n/9)) becomes arbitrarily large.
Therefore, the limit is 1.
In all three cases, the limits are equal to 1.