I have less than 24 hours to find the binomial probability of P(20<x<23)=

and P(x>23)= if n=30 p=.23 u=6.9 q=.77 and the standard deviation is 2.3 can someone please show me how to get the outcome they want?

To find the binomial probability for a given problem, we can use the formula:

P(x) = (nCx) * p^x * q^(n-x),

where P(x) is the probability of getting x successes in n trials, nCx is the number of combinations of n things taken x at a time, p is the probability of success in a single trial, and q is the probability of failure in a single trial (q = 1 - p).

For the first part of your question, P(20 < x < 23), we need to find the sum of probabilities for x = 21, 22, and 23, since it represents the probability of getting between 20 and 23 successes, inclusive. Let's calculate each individual probability step by step:

P(21) = (30C21) * (0.23^21) * (0.77^9)
P(22) = (30C22) * (0.23^22) * (0.77^8)
P(23) = (30C23) * (0.23^23) * (0.77^7)

Using the combination formula: nCr = n! / (r!(n-r)!)

Now, calculating the individual probabilities:

P(21) = (30C21) * (0.23^21) * (0.77^9)
= (30! / (21! * (30-21)!)) * (0.23^21) * (0.77^9)

Similarly, calculate P(22) and P(23) using the same formula. Finally, add the three probabilities together to find the desired probability:

P(20 < x < 23) = P(21) + P(22) + P(23)

For the second part of your question, P(x > 23), we need to calculate the probability of getting more than 23 successes, which is:

P(x > 23) = 1 - P(x ≤ 23)

To calculate P(x ≤ 23), we can calculate the sum of probabilities for x = 0 to 23 (inclusive) and subtract it from 1. We need to calculate individual probabilities for each of these x values using the formula discussed earlier, and then add them up.