Assume a binomial probability distribution has p = .60 and n = 200.

c. What is the probability of 100 to 110 successes (to 4 decimals)?


d. What is the probability of 130 or more successes (to 4 decimals)?

Here is an online normal distribution calculator:

http://davidmlane.com/hyperstat/z_table.html

by the way that calculator uses x - mean for z instead of (x-mean)/sigma

To find the probability in a binomial probability distribution, we can use the binomial probability formula:

P(x) = (nCx) * p^x * (1-p)^(n-x)

Where:
P(x) = probability of x successes
n = number of trials
x = number of successes
p = probability of success in a single trial
(1-p) = probability of failure in a single trial

In this case, we have:
n = 200 (number of trials)
p = 0.60 (probability of success)

c. To find the probability of 100 to 110 successes, we need to sum up the probabilities of all the individual outcomes between 100 and 110.

P(100 to 110) = P(100) + P(101) + ... + P(110)

We can calculate each individual probability using the binomial probability formula, and then sum them up.

d. To find the probability of 130 or more successes, we need to calculate the probabilities of 130, 131, ..., 200 and sum them up.

P(130 or more) = P(130) + P(131) + ... + P(200)

Now let's calculate these probabilities step by step.

c. Probability of 100 to 110 successes:
P(100 to 110) = P(100) + P(101) + ... + P(110)

To calculate each individual probability, we can plug in the values into the binomial probability formula:

P(x) = (nCx) * p^x * (1-p)^(n-x)

For example, let's calculate P(100):

P(100) = (200C100) * (0.60)^100 * (1-0.60)^(200-100)

To calculate (200C100), we can use the combination formula:

(200C100) = n! / (x! * (n-x)!)

(200C100) = 200! / (100! * (200-100)!)

(200C100) can be calculated using factorials.

Similarly, we can calculate P(101), P(102), ..., P(110).

After calculating each individual probability, we can sum them up to find the total probability.

d. Probability of 130 or more successes:
P(130 or more) = P(130) + P(131) + ... + P(200)

To calculate each individual probability, we can plug in the values into the binomial probability formula:

P(x) = (nCx) * p^x * (1-p)^(n-x)

For example, let's calculate P(130):

P(130) = (200C130) * (0.60)^130 * (1-0.60)^(200-130)

To calculate (200C130), we can use the combination formula:

(200C130) = n! / (x! * (n-x)!)

(200C130) can be calculated using factorials.

Similarly, we can calculate P(131), P(132), ..., P(200).

After calculating each individual probability, we can sum them up to find the total probability.

well, you have enough data points to use normal distribution rather than computing with binomials forever and a day. Otherwise use a program or spreadsheet to compute all the binomial coefs and powers and products

mean = n p = 200 * .6 = 120

sigma^2 = 120 (.4) = 48 so sigma = 6.93

for 100
z = (100 - 120)/6.93 = -20/6.93 = -2.89
for 110
z = (110-120)/6.93) = -1.45
so go to normal distribution table and find from -2.89 to -1.45
about .06

for 130
z =10/6.93 = +1.45
up to 1.45 in the table = .925
so about 1 - .925 = .075

I do not have an accurate table handy so you will have to do it more accurately.