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.
It is estimated that 3.7% of the general population will live past their 90th birthday. In a graduating class of 718 high school seniors, find the following probabilities. (Round your answers to four decimal places.)
A button hyperlink to the SALT program that reads: Use SALT.
(a) 15 or more will live beyond their 90th birthday
(b) 30 or more will live beyond their 90th birthday
(c) between 25 and 35 will live beyond their 90th birthday
(d) more than 40 will live beyond their 90th birthday
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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 distribution are satisfied. These conditions are:
1. The number of trials must be large: The number of high school seniors in this graduating class is 718, which is considered large enough.
2. The probability of success must be close to the same for each trial: The probability of a high school senior living past their 90th birthday is estimated to be 3.7%, which is approximately the same for each senior.
Given that these conditions are satisfied, we can proceed to use the normal distribution to estimate the requested probabilities.
(a) To find the probability that 15 or more will live beyond their 90th birthday, we can use the normal distribution with the mean (μ) and standard deviation (σ) of the binomial distribution. For a binomial distribution, the mean (μ) is equal to n*p, and the standard deviation (σ) is equal to sqrt(n*p*(1-p)).
μ = n*p = 718 * 0.037 = 26.566
σ = sqrt(n*p*(1-p)) = sqrt(718 * 0.037 * (1-0.037)) = 5.135
Using the normal distribution, we can calculate the probability:
P(X ≥ 15) = 1 - P(X < 15) = 1 - P(Z < (15 - μ)/σ) = 1 - P(Z < (15 - 26.566)/5.135)
Using a standard normal distribution table or a calculator, we find that P(Z < -2.243) is approximately 0.0122.
Therefore, P(X ≥ 15) = 1 - 0.0122 ≈ 0.9878 (rounded to four decimal places).
(b) To find the probability that 30 or more will live beyond their 90th birthday, we follow the same steps as in part (a).
μ = 26.566
σ = 5.135
P(X ≥ 30) = 1 - P(X < 30) = 1 - P(Z < (30 - 26.566)/5.135)
Using a standard normal distribution table or a calculator, we find that P(Z < 0.670) is approximately 0.7486.
Therefore, P(X ≥ 30) = 1 - 0.7486 ≈ 0.2514 (rounded to four decimal places).
(c) To find the probability that between 25 and 35 will live beyond their 90th birthday, we calculate the probabilities separately for both values and subtract the lower probability from the higher probability.
P(25 ≤ X ≤ 35) = P(X ≤ 35) - P(X < 25) = P(Z < (35 - μ)/σ) - P(Z < (25 - μ)/σ)
Using the same values for μ and σ as in previous parts:
P(25 ≤ X ≤ 35) = P(Z < (35 - 26.566)/5.135) - P(Z < (25 - 26.566)/5.135)
Using a standard normal distribution table or a calculator, we find that P(Z < 1.649) is approximately 0.9501 and P(Z < -0.2876) is approximately 0.3867.
Therefore, P(25 ≤ X ≤ 35) = 0.9501 - 0.3867 ≈ 0.5634 (rounded to four decimal places).
(d) To find the probability that more than 40 will live beyond their 90th birthday, we calculate the probability of the complement event (less than or equal to 40) and subtract it from 1.
P(X > 40) = 1 - P(X ≤ 40) = 1 - P(Z < (40 - μ)/σ)
Using the same values for μ and σ as in previous parts:
P(X > 40) = 1 - P(Z < (40 - 26.566)/5.135)
Using a standard normal distribution table or a calculator, we find that P(Z < 2.481) is approximately 0.9933.
Therefore, P(X > 40) = 1 - 0.9933 ≈ 0.0067 (rounded to four decimal places).
Keep in mind that these probabilities are estimated using the normal distribution as an approximation to the binomial distribution.
To determine whether it is appropriate to use the normal approximation to the binomial, we need to check if the conditions for using such an approximation are met. The conditions are:
1. The number of trials, n, is large.
2. The population proportion of success, p, is small.
3. The sample size, np, and nq are both greater than or equal to 10.
Let's go through each condition and see if they are satisfied in this problem.
1. The number of trials, n = 718, is large. This condition is satisfied.
2. The population proportion of success, p = 0.037, is small. This means that the probability of an individual living beyond their 90th birthday is small. This condition is satisfied.
3. We need to check if the sample size times the population proportion of success (np) and the sample size times the population proportion of failure (nq) are both greater than or equal to 10.
In this case, np = 718 * 0.037 ≈ 26.6 and nq = 718 * (1 - 0.037) ≈ 691.4. Both np and nq are greater than or equal to 10. This condition is satisfied.
Since all the conditions for using the normal approximation to the binomial are satisfied, we can proceed to use the normal distribution to estimate the requested probabilities.
To estimate the requested probabilities, we can use the normal approximation to the binomial by finding the mean (μ) and standard deviation (σ) of the distribution.
The mean (μ) of the binomial distribution is n * p, which is 718 * 0.037 ≈ 26.6.
The standard deviation (σ) of the binomial distribution is sqrt(n * p * q), which is sqrt(718 * 0.037 * (1 - 0.037)) ≈ 5.14.
Now we can use the normal distribution with mean 26.6 and standard deviation 5.14 to estimate the requested probabilities.
(a) To find the probability that 15 or more will live beyond their 90th birthday, we can use the normal distribution by converting 15 to a standardized z-score and finding the area under the normal curve to the right of that z-score.
The z-score can be calculated using the formula z = (x - μ) / σ, where x = 15, μ = 26.6, and σ = 5.14.
Substituting the values, we get z = (15 - 26.6) / 5.14 ≈ -2.27.
To find the probability, we can use a z-table or a calculator to find the area to the right of -2.27. The probability is approximately 0.9886.
(b) To find the probability that 30 or more will live beyond their 90th birthday, we can use the same approach as in part (a).
Calculate the z-score: z = (30 - 26.6) / 5.14 ≈ 0.66.
Using a z-table or a calculator, find the area to the right of 0.66. The probability is approximately 0.2546.
(c) To find the probability that between 25 and 35 will live beyond their 90th birthday, we need to find the probabilities for both 25 and 35 separately and then subtract the two probabilities.
Calculate the z-scores:
For 25, z = (25 - 26.6) / 5.14 ≈ -0.31.
For 35, z = (35 - 26.6) / 5.14 ≈ 1.62.
Using a z-table or a calculator, find the area to the right of -0.31 and the area to the right of 1.62. Subtract the two probabilities to estimate the probability between 25 and 35. The answer is approximately 0.4386.
(d) To find the probability that more than 40 will live beyond their 90th birthday, we can calculate the z-score for 40 and find the area to the right of that z-score.
Calculate the z-score: z = (40 - 26.6) / 5.14 ≈ 2.61.
Using a z-table or a calculator, find the area to the right of 2.61. The probability is approximately 0.0047.
Remember to round your answers to four decimal places, as specified in the problem.
These are the estimated probabilities using the normal distribution to approximate the binomial distribution.