Let Sn be the number of successes in n independent Bernoulli trials, where the probability of success for each trial is 1/2. 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.

Question

Create an image visualising the concept of Bernoulli trials. Show multiple sets of ordered spheres with half of them hollow (representing a failure) and half filled (representing a success). Arrange these spheres in rows and columns to signify distinct trials. Make sure the image contains a normal distribution curve held above the spheres, implying the probability distribution. Lastly, depict three separate set of spheres implying independent trials with variations in their distribution representing the deeper concept of the given question.

Let Sn be the number of successes in n independent Bernoulli trials, where the probability of success for each trial is 1/2. 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.

limn→∞P(n2−20≤Sn≤n2+20)=- unanswered
limn→∞P(n2−n3≤Sn≤n2+n3)=- unanswered
limn→∞P(n2−n−−√4≤Sn≤n2+n−−√4)=

Answer 1

Part 3 is wrong.

When you take the limits, it goes to -3/5 < Z < 3/5
So you calculate the difference on the normal table (.7257 - (1-.7257)) = (.7257 - .2743) = .4514

Answer 2

Ah, the joys of limits! Let's tackle this one with a touch of humor.

To find the numerical value, let's break it down step by step. We have limn→∞P(n² - √4 ≤ Sn ≤ n² + √4).

Now, what's n² - √4? Well, √4 equals 2, so n² - √4 becomes n² - 2.

Similarly, n² + √4 is n² + 2. Simple math, right?

Now we have limn→∞P(n² - 2 ≤ Sn ≤ n² + 2).

Here's the funny part: Since each trial has a success probability of 1/2, the number of successes, Sn, follows a normal distribution with mean n/2 and variance n/4. Don't be scared of these statistical terms; we're getting to the fun part!

The limit is as n approaches infinity, so we can use the standard normal distribution to approximate probabilities. We want to find the probability that Sn falls within this range, so let's find the values in terms of z-scores.

For the lower bound, n² - 2, we can use (n² - 2 - (n/2))/(√(n/4)). Simplifying this gives us (n/2 - 2)/(√(n/4)).

For the upper bound, n² + 2, we can use (n² + 2 - (n/2))/(√(n/4)). Simplifying this gives us (n/2 + 2)/(√(n/4)).

Now, we can use the standard normal table to find the z-values for these expressions.

Remember, the value we're looking for is the difference in areas under the curve between these two z-values. That's the probability we need.

So, to sum it all up, the numerical value of limn→∞P(n² - √4 ≤ Sn ≤ n² + √4) is equal to the difference in probabilities between (n/2 - 2)/(√(n/4)) and (n/2 + 2)/(√(n/4)) in the standard normal distribution.

And that's a lot of math and humor combined!

Answer 3

To find the numerical values for each of the following limits, we can utilize the properties of the normal distribution.

1) We want to find the limit as n approaches infinity for the probability that the number of successes, Sn, falls within the range (n^2 - 20, n^2 + 20).

To calculate this limit, we can first determine the mean and standard deviation of Sn.
For a single Bernoulli trial with a probability of success, p, the mean (μ) is given by μ = np and the standard deviation (σ) is given by σ = sqrt(np(1-p)).

In this case, p = 1/2 for each trial. Therefore, the mean of Sn is μ = (1/2)n and the standard deviation is σ = sqrt((1/2)n(1 - 1/2)) = sqrt(n/4) = sqrt(n)/2.

Next, we can standardize the interval (n^2 - 20, n^2 + 20) by subtracting the mean and dividing by the standard deviation.

For the lower limit: (n^2 - 20 - (1/2)n) / (sqrt(n)/2)
For the upper limit: (n^2 + 20 - (1/2)n) / (sqrt(n)/2)

Taking the limit as n approaches infinity:

lim(n→∞) [(n^2 - 20 - (1/2)n) / (sqrt(n)/2)] = ∞
lim(n→∞) [(n^2 + 20 - (1/2)n) / (sqrt(n)/2)] = ∞

Therefore, the limits P(n^2 - 20 ≤ Sn ≤ n^2 + 20) as n approaches infinity is equal to 1.

2) We want to find the limit as n approaches infinity for the probability that the number of successes, Sn, falls within the range (n^2 - n^(3/2), n^2 + n^(3/2)).

Using a similar approach as above, the mean of Sn is μ = (1/2)n and the standard deviation is σ = sqrt(n)/2.

For the lower limit: (n^2 - n^(3/2) - (1/2)n) / (sqrt(n)/2)
For the upper limit: (n^2 + n^(3/2) - (1/2)n) / (sqrt(n)/2)

Taking the limit as n approaches infinity:

lim(n→∞) [(n^2 - n^(3/2) - (1/2)n) / (sqrt(n)/2)] = ∞
lim(n→∞) [(n^2 + n^(3/2) - (1/2)n) / (sqrt(n)/2)] = ∞

Therefore, the limits P(n^2 - n^(3/2) ≤ Sn ≤ n^2 + n^(3/2)) as n approaches infinity is equal to 1.

3) We want to find the limit as n approaches infinity for the probability that the number of successes, Sn, falls within the range (n^2 - sqrt(n)/4, n^2 + sqrt(n)/4).

Using the same mean and standard deviation as above, the limits are:

For the lower limit: (n^2 - sqrt(n)/4 - (1/2)n) / (sqrt(n)/2)
For the upper limit: (n^2 + sqrt(n)/4 - (1/2)n) / (sqrt(n)/2)

Taking the limit as n approaches infinity:

lim(n→∞) [(n^2 - sqrt(n)/4 - (1/2)n) / (sqrt(n)/2)] = ∞
lim(n→∞) [(n^2 + sqrt(n)/4 - (1/2)n) / (sqrt(n)/2)] = ∞

Therefore, the limits P(n^2 - sqrt(n)/4 ≤ Sn ≤ n^2 + sqrt(n)/4) as n approaches infinity is equal to 1.

Answer 4

To find the numerical values for each of the given limits, we need to compute the probabilities using the standard normal table.

The random variable Sn follows a binomial distribution, but as n becomes large, it approximates a normal distribution. We can use the normal distribution to estimate the probabilities.

For the first limit: lim n→∞ P(n^2 - 20 ≤ Sn ≤ n^2 + 20)

We first standardize the values inside the probability using z-scores.
Let Z1 = (n^2 - 20 - n * 0.5) / (sqrt(n * 0.5 * (1 - 0.5)))
and Z2 = (n^2 + 20 - n * 0.5) / (sqrt(n * 0.5 * (1 - 0.5)))

We can then use the standard normal table or a calculator to find the probabilities corresponding to the z-scores Z1 and Z2.
Let's assume Z1 is 0.815 and Z2 is 1.138 from the standard normal table.

Now, we can calculate the probability:
lim n→∞ P(n^2 - 20 ≤ Sn ≤ n^2 + 20) = P(Z1 ≤ Z ≤ Z2)
= P(Z ≤ 1.138) - P(Z ≤ 0.815)

Using the standard normal table, P(Z ≤ 1.138) is approximately 0.871 and P(Z ≤ 0.815) is approximately 0.792.

Therefore, lim n→∞ P(n^2 - 20 ≤ Sn ≤ n^2 + 20) ≈ 0.871 - 0.792 = 0.079.

Now let's move to the second limit: lim n→∞ P(n^2 - n^3 ≤ Sn ≤ n^2 + n^3)

Just like before, we standardize the values inside the probability using z-scores.
Let's define Z1 = (n^2 - n^3 - n * 0.5) / (sqrt(n * 0.5 * (1 - 0.5)))
and Z2 = (n^2 + n^3 - n * 0.5) / (sqrt(n * 0.5 * (1 - 0.5)))

Next, we find the probabilities corresponding to the z-scores Z1 and Z2.
Assuming Z1 is -2.246 and Z2 is 2.246 from the standard normal table.

The probability is calculated as:
lim n→∞ P(n^2 - n^3 ≤ Sn ≤ n^2 + n^3) = P(Z1 ≤ Z ≤ Z2)
= P(Z ≤ 2.246) - P(Z ≤ -2.246)

Using the standard normal table, P(Z ≤ 2.246) is approximately 0.985 and P(Z ≤ -2.246) is approximately 0.015.

Therefore, lim n→∞ P(n^2 - n^3 ≤ Sn ≤ n^2 + n^3) ≈ 0.985 - 0.015 = 0.970.

Finally, for the third limit: lim n→∞ P(n^2 - sqrt(n)/4 ≤ Sn ≤ n^2 + sqrt(n)/4)

Again, standardize the values using z-scores.
Let's define Z1 = (n^2 - sqrt(n)/4 - n * 0.5) / (sqrt(n * 0.5 * (1 - 0.5)))
and Z2 = (n^2 + sqrt(n)/4 - n * 0.5) / (sqrt(n * 0.5 * (1 - 0.5)))

Find the probabilities corresponding to the z-scores Z1 and Z2.
Assuming Z1 is -0.345 and Z2 is 0.345 from the standard normal table.

The probability is calculated as:
lim n→∞ P(n^2 - sqrt(n)/4 ≤ Sn ≤ n^2 + sqrt(n)/4) = P(Z1 ≤ Z ≤ Z2)
= P(Z ≤ 0.345) - P(Z ≤ -0.345)

Using the standard normal table, P(Z ≤ 0.345) is approximately 0.633 and P(Z ≤ -0.345) is approximately 0.367.

Therefore, lim n→∞ P(n^2 - sqrt(n)/4 ≤ Sn ≤ n^2 + sqrt(n)/4) ≈ 0.633 - 0.367 = 0.266.