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 87% 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 212 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 212 children taken now, what is the probability that 50 or more have high blood-lead levels? (Round your answer to three decimal places.)
Very frustrating- found the np, nq, mean, standard deviation, z, and the result is too big to find on the number chart.
(a)
p=0.87, q=0.13, n=212
need P(X>=50)
mean = np = 212*0.87 = 184.44
variance = npq = 23.9772
σ = sqrt(variance) = 4.8967
To find
P(X>=50)=P(Z>(50-184.44)/4.8967)
=P(Z>-27.45)=1.000 from Normal probability table.
Do a similar calculation for (b) using p=0.13
To determine if it is appropriate to use the normal approximation to the binomial distribution in this case, we need to check if the conditions for the normal approximation are satisfied. These conditions are:
1) The sample size is large enough: According to the Central Limit Theorem, the sample size should ideally be greater than or equal to 30 to use the normal approximation. In both parts (a) and (b), the sample size is 212, so this condition is satisfied.
2) The number of successes and failures in the sample are at least 10: We need to check if both np and n(1-p) is greater than or equal to 10, where n is the sample size and p is the probability of success.
(a) For the sample taken more than a decade ago:
np = 212 * 0.87 = 184.44
nq = 212 * (1 - 0.87) = 27.56
Both np and nq are greater than 10, so this condition is satisfied.
(b) For the sample taken now:
np = 212 * 0.13 = 27.56
nq = 212 * (1 - 0.13) = 184.44
Both np and nq are greater than 10, so this condition is also satisfied.
Since both conditions for using the normal approximation are satisfied in both parts (a) and (b), we can proceed with estimating the requested probabilities using the normal distribution.
To find the probabilities using the normal distribution, we can calculate the z-scores for each scenario and use the standard normal distribution table or calculator.
(a) For the sample taken more than a decade ago:
Mean (μ) = np = 212 * 0.87 = 184.44
Standard deviation (σ) = sqrt(npq) = sqrt(212 * 0.87 * 0.13) ≈ 5.302
To find the probability that 50 or more children had high blood-lead levels, we need to find P(X ≥ 50), where X follows a normal distribution with mean μ = 184.44 and standard deviation σ ≈ 5.302.
Let's calculate the z-score for X = 50:
z = (X - μ) / σ = (50 - 184.44) / 5.302 ≈ -26.832
Now we can look up the probability corresponding to this z-score in the standard normal distribution table or use a calculator.
(b) For the sample taken now:
Mean (μ) = np = 212 * 0.13 = 27.56
Standard deviation (σ) = sqrt(npq) = sqrt(212 * 0.13 * 0.87) ≈ 5.302
To find the probability that 50 or more children have high blood-lead levels, we need to find P(X ≥ 50), where X follows a normal distribution with mean μ = 27.56 and standard deviation σ ≈ 5.302.
Let's calculate the z-score for X = 50:
z = (X - μ) / σ ≈ (50 - 27.56) / 5.302 ≈ 4.25
Now we can look up the probability corresponding to this z-score in the standard normal distribution table or use a calculator.
Apologies for any frustration caused by a result being too big to find on a number chart. In situations like these, using the z-scores to find probabilities from the standard normal distribution table or using a calculator is usually the best way to obtain accurate results.
To check if it is appropriate to use the normal approximation to the binomial in this problem, we need to ensure that both np and nq are greater than or equal to 10.
First, let's calculate np and nq for the original scenario (more than a decade ago):
np = n * p = 212 * 0.87 = 184.44
nq = n * (1 - p) = 212 * (1 - 0.87) = 27.56
Since both np and nq are greater than 10, we can use the normal approximation to the binomial distribution.
Next, let's calculate np and nq for the current scenario (now):
np = n * p = 212 * 0.13 = 27.56
nq = n * (1 - p) = 212 * (1 - 0.13) = 184.44
Again, both np and nq are greater than 10, so we can use the normal approximation here as well.
Now, let's calculate the probabilities using the normal distribution.
(a) In a random sample of 212 children taken more than a decade ago, we want to find the probability that 50 or more had high blood-lead levels.
To solve this, we can approximate the binomial distribution with a normal distribution using the mean (mu) and standard deviation (sigma) derived from the binomial distribution.
mu = np = 184.44
sigma = √(npq) = √(27.56 * 0.87) = 4.349
Now we can calculate the z-score using the formula:
z = (x - mu) / sigma
where x is the number of children with high blood-lead levels.
For x = 50:
z = (50 - 184.44) / 4.349 = -61.69
Using a standard normal distribution table or a calculator, we can find the probability associated with this z-score. However, since the z-value is too large to find on the number chart, we can use a calculator or statistical software to find the probability directly. The result is approximately 0.
Therefore, the probability that 50 or more children had high blood-lead levels more than a decade ago is very close to 0.
(b) In a random sample of 212 children taken now, we want to find the probability that 50 or more have high blood-lead levels.
Using the same approach as before, we find:
mu = np = 27.56
sigma = √(npq) = √(27.56 * 0.13) = 3.04
For x = 50:
z = (50 - 27.56) / 3.04 = 7.39
Again, using a calculator or statistical software, we can find the probability associated with this z-score. However, this time the result falls into a range that can be found on the number chart. You can look up the probability associated with the z-score of approximately 7.39 and round it to three decimal places.
Note: The result will be a very small number close to 1.000.
To summarize:
(a) The probability that 50 or more children had high blood-lead levels more than a decade ago is very close to 0.
(b) The probability that 50 or more children have high blood-lead levels now is very close to 1.000