In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities.
More than a decade ago, high levels of lead in the blood put 89% of children at risk. A concerted effort was made to remove lead from the environment. Now, suppose only 13% of children in the United States are at risk of high blood-lead levels.
(a) In a random sample of 204 children taken more than a decade ago, what is the probability that 50 or more had high blood-lead levels? (Round your answer to three decimal places.)
(b) In a random sample of 204 children taken now, what is the probability that 50 or more have high blood-lead levels? (Round your answer to three decimal places.)
a. Mu = np = .89(204) = 181.56
p(r => 49.5)
sd = 4.469
z = (49.5-181.56)/4.469 =
b
mu = .13(204) = 26.52
Sd = 4.803
z = (49.5-26.52)/4.80 = 4.7875
In order to determine whether it is appropriate to use the normal approximation to the binomial, we need to check if the conditions for using the normal approximation are satisfied. These conditions are as follows:
1. The sample size should be large enough: For the normal approximation to be appropriate, both the number of successes (np) and the number of failures (nq) should be at least 10. Let's check this condition for each part of the problem.
(a) For the sample taken more than a decade ago: The sample size is 204, and the probability of success is 0.89. Thus, np = 204 * 0.89 = 181.56 is greater than 10, and nq = 204 * (1 - 0.89) = 22.44 is also greater than 10.
(b) For the current sample: The sample size is still 204, but the probability of success is now 0.13. Thus, np = 204 * 0.13 = 26.52 is greater than 10, and nq = 204 * (1 - 0.13) = 177.48 is also greater than 10.
2. The observations should be independent: In order to use the normal approximation, the 204 children in each sample need to be randomly selected and their blood-lead levels should be independent of each other. We assume this is the case in this problem.
Since both conditions are satisfied for both parts of the problem, we can use the normal distribution to estimate the requested probabilities.
(a) Probability that 50 or more children had high blood-lead levels more than a decade ago:
To solve this, we convert the binomial distribution to a normal distribution using the mean and standard deviation of the binomial distribution.
The mean (μ) of the binomial distribution is equal to n * p = 204 * 0.89 = 181.56.
The standard deviation (σ) of the binomial distribution is equal to the square root of (n * p * q) = sqrt(204 * 0.89 * (1 - 0.89)) ≈ 7.387.
Now, with the normal distribution, we need to calculate the probability of getting a value of 50 or more, which is equivalent to finding P(X >= 50) using the normal approximation.
Using the mean (μ = 181.56) and standard deviation (σ = 7.387), we can standardize the variable X = 50 and calculate the z-score:
z = (X - μ) / σ = (50 - 181.56) / 7.387 = -17.93.
We then find the probability from the standard normal distribution table associated with z = -17.93, which is essentially 0. The probability that 50 or more children had high blood-lead levels more than a decade ago is approximately 0.000.
(b) Probability that 50 or more children have high blood-lead levels now:
Using the same approach as in part (a), we compute the mean (μ) and standard deviation (σ) of the binomial distribution:
μ = n * p = 204 * 0.13 = 26.52.
σ = sqrt(n * p * q) = sqrt(204 * 0.13 * (1 - 0.13)) ≈ 6.615.
Next, we standardize the variable X = 50 and calculate the z-score:
z = (X - μ) / σ = (50 - 26.52) / 6.615 = 3.55.
We then find the probability from the standard normal distribution table associated with z = 3.55, which is approximately 0.9998.
The probability that 50 or more children have high blood-lead levels now is approximately 0.9998.
To determine whether it is appropriate to use the normal approximation to the binomial distribution in this problem, we need to check if the conditions for using the normal approximation are satisfied. The conditions are:
1. The number of trials (n) is large enough: np ≥ 10 and n(1-p) ≥ 10.
2. The observations are independent.
Now let's apply these conditions to the given problem:
(a) In a random sample of 204 children taken more than a decade ago, we are looking for the probability that 50 or more had high blood-lead levels. The probability of a child having high blood-lead levels is 0.89.
To check the conditions:
1. np = 204 * 0.89 = 181.56 ≥ 10, and n(1-p) = 204 * (1 - 0.89) = 22.44 ≥ 10.
Both conditions are satisfied, so we can proceed with using the normal approximation to the binomial distribution.
To estimate the probability, we can use the normal distribution with mean (μ) = n * p and standard deviation (σ) = √(n * p * (1 - p)).
μ = 204 * 0.89 = 181.56
σ = √(204 * 0.89 * (1 - 0.89)) ≈ 6.85
We are interested in the probability that 50 or more children had high blood-lead levels, which can be expressed as P(X ≥ 50), where X follows a normal distribution with mean μ and standard deviation σ.
To find this probability, we need to standardize the distribution to a standard normal distribution using the formula z = (x - μ) / σ.
For P(X ≥ 50), we can find the corresponding z-score:
z = (50 - 181.56) / 6.85 ≈ -22.34
Using a standard normal distribution table or a calculator, we find that the probability corresponding to z = -22.34 is practically zero. Therefore, the probability that 50 or more children had high blood-lead levels is very close to zero.
(b) In a random sample of 204 children taken now, we are looking for the probability that 50 or more have high blood-lead levels. The probability of a child having high blood-lead levels is now 0.13.
To check the conditions:
1. np = 204 * 0.13 = 26.52 ≥ 10, and n(1-p) = 204 * (1 - 0.13) = 177.48 ≥ 10.
Both conditions are satisfied, so we can use the normal approximation to the binomial distribution.
To estimate the probability, we can use the normal distribution with mean (μ) = n * p and standard deviation (σ) = √(n * p * (1 - p)).
μ = 204 * 0.13 = 26.52
σ = √(204 * 0.13 * (1 - 0.13)) ≈ 6.51
We are interested in the probability that 50 or more children have high blood-lead levels, which can be expressed as P(X ≥ 50), where X follows a normal distribution with mean μ and standard deviation σ.
To find this probability, we need to standardize the distribution to a standard normal distribution using the formula z = (x - μ) / σ.
For P(X ≥ 50), we can find the corresponding z-score:
z = (50 - 26.52) / 6.51 ≈ 3.62
Using a standard normal distribution table or a calculator, we find that the probability corresponding to z = 3.62 is approximately 0.0001. Therefore, the estimated probability that 50 or more children have high blood-lead levels is approximately 0.0001.