Find the limits
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.
Normal Table
1) limn→∞P(n/3 − 10 ≤ Sn ≤ n/3 + 10)=
unanswered
2) limn→∞P(n/3 − n/6 ≤ Sn ≤ n/3 + n/6)=
unanswered
3)limn→∞P(n/3 − √2n/5 ≤ Sn ≤ n/3 + √2n/5)=
unanswered
See Similar Questions below.
To find the limits for each of the given expressions, we can use the Central Limit Theorem.
According to the Central Limit Theorem, if Sn is the number of successes in n independent Bernoulli trials with probability of success p=1/3, then the distribution of Sn is approximately normal with mean μ = np and standard deviation σ = sqrt(np(1-p)).
1) limn→∞ P(n/3 − 10 ≤ Sn ≤ n/3 + 10):
Using the formula for the mean and standard deviation, we have μ = n(1/3) = n/3 and σ = sqrt(n(1/3)(2/3)) = sqrt(2n/9).
We want to find the probability that the number of successes falls within the range n/3 - 10 to n/3 + 10.
Using the standardized normal distribution, we can convert this range into z-scores:
z1 = (n/3 - 10 - μ)/σ = (n/3 - 10 - n/3)/(sqrt(2n/9)) = -10/sqrt(2n/9) = -10sqrt(9/2n)
z2 = (n/3 + 10 - μ)/σ = (n/3 + 10 - n/3)/(sqrt(2n/9)) = 10/sqrt(2n/9) = 10sqrt(9/2n)
As n approaches infinity, both z1 and z2 approach negative and positive infinity respectively. Therefore, the probability approaches 1.
So, limn→∞ P(n/3 − 10 ≤ Sn ≤ n/3 + 10) = 1.
2) limn→∞ P(n/3 − n/6 ≤ Sn ≤ n/3 + n/6):
Similarly, using the mean and standard deviation formulas, we have μ = n/3 and σ = sqrt(2n/9).
We want to find the probability that the number of successes falls within the range n/3 - n/6 to n/3 + n/6.
Using the standardized normal distribution, we can convert this range into z-scores:
z1 = (n/3 - n/6 - μ)/σ = (n/6)/(sqrt(2n/9)) = sqrt(3/2n)
z2 = (n/3 + n/6 - μ)/σ = (n/2)/(sqrt(2n/9)) = sqrt(9/2n)
As n approaches infinity, both z1 and z2 approach zero. Therefore, the probability approaches 0.5.
So, limn→∞ P(n/3 − n/6 ≤ Sn ≤ n/3 + n/6) = 0.5.
3) limn→∞ P(n/3 − √2n/5 ≤ Sn ≤ n/3 + √2n/5):
Using the mean and standard deviation formulas, we have μ = n/3 and σ = sqrt(2n/9).
We want to find the probability that the number of successes falls within the range n/3 - √2n/5 to n/3 + √2n/5.
Using the standardized normal distribution, we can convert this range into z-scores:
z1 = (n/3 - √2n/5 - μ)/σ = (-(√2n)/(3sqrt(5)))/(sqrt(2n/9)) = -√2/3√5
z2 = (n/3 + √2n/5 - μ)/σ = ((√2n)/(3sqrt(5)))/(sqrt(2n/9)) = √2/3√5
These z-scores remain constant as n approaches infinity.
By looking up these z-scores in the standard normal table, we can find that P(-√2/3√5 ≤ z ≤ √2/3√5) is approximately 0.281 (to three decimal places).
So, limn→∞ P(n/3 − √2n/5 ≤ Sn ≤ n/3 + √2n/5) = 0.281.
To find the limits of the probabilities given in each question, we can make use of the Central Limit Theorem. The Central Limit Theorem states that the distribution of the sum (or average) of a large number of independent and identically distributed random variables approaches a normal distribution, regardless of the shape of the original distribution. In this case, we have a sum of independent Bernoulli random variables, which satisfy the conditions for the Central Limit Theorem.
To apply the Central Limit Theorem, we need to calculate the mean and standard deviation of the distribution. In a Bernoulli trial with probability of success p, the mean is given by μ = np and the standard deviation is given by σ = √(np(1-p)).
In this case, we have p = 1/3 (probability of success at each trial) and n going to infinity.
1) For the first limit, we have Sn = n/3. The mean of Sn is μ = np = (n/3) * (1/3) = n/9. The standard deviation of Sn is σ = √(np(1-p)) = √((n/3)(2/3)) = √(2n/9).
To find limn→∞P(n/3 − 10 ≤ Sn ≤ n/3 + 10), we need to standardize the values n/3 - 10 and n/3 + 10 using the z-score formula.
Z1 = (n/3 - 10 - (n/9)) / (√(2n/9)) = -√(9/2)/3
Z2 = (n/3 + 10 - (n/9)) / (√(2n/9)) = √(9/2)/3
Now we can use the standard normal table to find the probability between these z-scores.
2) For the second limit, we have Sn = n/3. The mean of Sn is still μ = n/9. The standard deviation of Sn is still σ = √(2n/9).
To find limn→∞P(n/3 - n/6 ≤ Sn ≤ n/3 + n/6), we follow the same steps as above and standardize the values n/3 - n/6 and n/3 + n/6 using the z-score formula.
Z1 = (n/3 - n/6 - (n/9)) / (√(2n/9)) = -√(1/2)
Z2 = (n/3 + n/6 - (n/9)) / (√(2n/9)) = √(1/2)
Again, we can use the standard normal table to find the probability between these z-scores.
3) For the third limit, we have Sn = n/3. The mean of Sn is still μ = n/9. The standard deviation of Sn is still σ = √(2n/9).
To find limn→∞P(n/3 - √2n/5 ≤ Sn ≤ n/3 + √2n/5), we standardize the values n/3 - √2n/5 and n/3 + √2n/5 using the z-score formula.
Z1 = (n/3 - √2n/5 - (n/9)) / (√(2n/9)) = -√(9/10)
Z2 = (n/3 + √2n/5 - (n/9)) / (√(2n/9)) = √(9/10)
Once again, we can use the standard normal table to find the probability between these z-scores.
By referring to the standard normal table and finding the probabilities for the corresponding z-scores, we can provide numerical values for each of the limits requested in the questions.