find the normal approximation for the binomial probability that x=5, where n=12, and p=0.7. compare this probability to the value of p(x=5).

* the only thing that i have come up with is that value of p(x=5) is o.29 when looking it up on the table. how would i successfully complete this problem? please help.

Find the normal approximation for the binomial probability that x = 5, where n = 12 and p = 0.7. Compare this probability to the value of P(x=5) found in Table 2 of Appendix B in your textbook.

To find the normal approximation for the binomial probability that x=5, where n=12 and p=0.7, you can use the formula for the mean (μ) and standard deviation (σ) of the binomial distribution:

μ = n * p
σ = sqrt(n * p * (1-p))

In this case:
μ = 12 * 0.7 = 8.4
σ = sqrt(12 * 0.7 * (1-0.7)) = sqrt(2.52) ≈ 1.59

To calculate the normal approximation, we need to standardize the value of x=5 using the z-score formula:

z = (x - μ) / σ

For x=5:
z = (5 - 8.4) / 1.59 ≈ -2.14

Next, we can use a standard normal distribution table or software to find the probability associated with the z-score of -2.14. The cumulative probability for a z-score of -2.14 is approximately 0.0163.

Therefore, the normal approximation for the binomial probability that x=5 is approximately 0.0163.

To compare this with the exact binomial probability, you mentioned that p(x=5) is 0.29. Please double-check this value as it seems to be different from the normal approximation result. If the correct value is indeed 0.29, then there is a significant difference between the normal approximation and the exact binomial probability. In such cases, it is recommended to use the exact binomial probability rather than relying on the approximation.