find the normal appproximation for the binomial probabiolity the x=5,where n=12, and p= 0.7. compare this to the value of p(x=5).
*please show calculations and work. want to compare my calculations. thank you.
mary
To find the normal approximation for the binomial probability, we can use the formula:
P(x) = (nCx) * p^x * (1-p)^(n-x)
where n is the number of trials, x is the number of successes, p is the probability of success, and (nCx) is the binomial coefficient, given by n! / (x! * (n-x)!).
In this case, we are interested in finding P(x=5) when n=12 and p=0.7.
First, let's calculate (nCx):
(nC5) = 12! / (5! * (12-5)!) = 12! / (5! * 7!) = (12*11*10*9*8) / (5*4*3*2*1) = 792
Now, let's calculate P(x=5):
P(5) = (nC5) * p^5 * (1-p)^(n-5) = 792 * (0.7)^5 * (1-0.7)^(12-5)
P(5) = 792 * (0.7)^5 * (0.3)^7 ≈ 0.0996 (rounded to four decimal places)
Next, to find the normal approximation, we need to calculate the mean (μ) and the standard deviation (σ) using the formulas:
μ = n * p
σ = sqrt(n * p * (1-p))
For this case, μ = 12 * 0.7 = 8.4 and σ = sqrt(12 * 0.7 * 0.3) ≈ 1.8723 (rounded to four decimal places).
Finally, we can use these values to find the normal approximation for P(x=5) using the Z-score formula:
Z = (x - μ) / σ
Z = (5 - 8.4) / 1.8723 ≈ -1.8127 (rounded to four decimal places).
Now, we can use a Z-table or a calculator to find the probability associated with Z = -1.8127. Let's assume the value is P(Z ≤ -1.8127) ≈ 0.0343 (rounded to four decimal places).
Comparing this to the actual binomial probability, we can see that the normal approximation is close but not exact. The actual probability P(x=5) was approximately 0.0996, while the normal approximation gave us a probability of approximately 0.0343. So, the normal approximation underestimates the true binomial probability in this case.