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.
1. lim P (n/2 - 20 < Sn < n/2 + 20)
2. lim P (n/2 - n/3 < Sn < n/2 + n/3)
3. lim P (n/2 - sqrt(n)/4 < Sn < n/2 + sqrt(n)/4)
1. -> 0.000
2. -> 1.000
3. -> 0.383
.5
1
.6286
To find the numerical values for each of the given limits, we need to use the Normal approximation to the Binomial distribution. This approximation is valid when n is large and the probability of success (p) is not close to 0 or 1.
The mean (μ) and standard deviation (σ) of a Binomial distribution are given by:
μ = n * p
σ = sqrt(n * p * (1 - p))
In this case, p = 1/2 since the probability of success for each trial is 1/2.
Now, we can determine the limits and compute the probabilities using the Normal distribution.
1. lim P(n/2 - 20 < Sn < n/2 + 20):
Using the Normal approximation, we approximate Sn as a Normally distributed random variable with mean μ = n * p = n * (1/2) = n/2 and standard deviation σ = sqrt(n * p * (1 - p)) = sqrt(n * (1/2) * (1 - 1/2)) = sqrt(n/4) = sqrt(n)/2.
To find the probability between the limits, we can use the standard Normal distribution by standardizing the limits:
Z1 = (n/2 - 20 - μ) / σ = (n/2 - 20 - n/2) / (sqrt(n)/2) = -40 / (sqrt(n)/2) = -80 / sqrt(n)
Z2 = (n/2 + 20 - μ) / σ = (n/2 + 20 - n/2) / (sqrt(n)/2) = 40 / (sqrt(n)/2) = 80 / sqrt(n)
Now, we can calculate the probability:
P(n/2 - 20 < Sn < n/2 + 20) = P(Z1 < Z < Z2) = P(Z < Z2) - P(Z < Z1), where Z is a standard Normal random variable.
Using a standard Normal distribution table or a statistical software, we can find the probabilities corresponding to Z1 and Z2 and subtract them to get the final result.
2. lim P(n/2 - n/3 < Sn < n/2 + n/3):
Using the same approach as above, we find the mean and standard deviation for the distribution:
μ = n/2
σ = sqrt(n/4) = sqrt(n)/2
Again, we standardize the limits by dividing by the standard deviation:
Z1 = (n/2 - n/3 - μ) / σ
Z2 = (n/2 + n/3 - μ) / σ
Then, we calculate the probability:
P(n/2 - n/3 < Sn < n/2 + n/3) = P(Z1 < Z < Z2) = P(Z < Z2) - P(Z < Z1).
3. lim P(n/2 - sqrt(n)/4 < Sn < n/2 + sqrt(n)/4):
Similarly, we calculate the mean and standard deviation:
μ = n/2
σ = sqrt(n)/2
Standardizing the limits:
Z1 = (n/2 - sqrt(n)/4 - μ) / σ
Z2 = (n/2 + sqrt(n)/4 - μ) / σ
Then, we compute the probability:
P(n/2 - sqrt(n)/4 < Sn < n/2 + sqrt(n)/4) = P(Z1 < Z < Z2) = P(Z < Z2) - P(Z < Z1).
Remember to use a standard Normal distribution table or a statistical software to find the corresponding probabilities for the Z-values and calculate the final result.