np >= 5 and nq >= 5 , estimate P(at least 11) with n = 13 and p = 0.6 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(at least 11) using the normal distribution as an approximation to the binomial distribution, we need to check if the conditions for the normal approximation are satisfied:
1. np ≥ 5 and nq ≥ 5.
Here, n is the number of trials (13) and p is the probability of success on a single trial (0.6). Let's calculate np and nq:
np = 13 * 0.6 = 7.8
nq = 13 * (1 - 0.6) = 5.2
Since both np and nq are greater than or equal to 5, the conditions for the normal approximation are satisfied.
We can now proceed with estimating P(at least 11) using the normal approximation.
We know that the binomial distribution can be approximated by a normal distribution with mean μ = np and standard deviation σ = √(npq). In this case, μ = 7.8 and σ ≈ √(7.8 * 0.4) ≈ 1.80.
To find the probability of at least 11 successes, we need to calculate the area under the normal curve to the right of 10.5 (since we want at least 11).
Using z-scores, we can standardize this value as follows:
z = (x - μ) / σ
where x = 10.5, μ = 7.8, and σ = 1.80.
z = (10.5 - 7.8) / 1.80 ≈ 1.50
Now, we can use a standard normal table or calculator to find the probability associated with a z-score of 1.50. Let's assume the corresponding area is denoted as P(Z > 1.50).
The probability of at least 11 successes, P(at least 11), can be calculated as:
P(at least 11) ≈ P(Z > 1.50)
Note: P(Z > 1.50) represents the cumulative probability from the z-score of 1.50 to positive infinity (the right-tail probability).
Using a standard normal table or calculator, you can find the value of P(Z > 1.50) or the equivalent value of P(Z ≤ -1.50) (since the normal distribution is symmetrical).