Ask Question
19 September, 17:31

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 745 high school seniors, find the following probabilities. (Round your answers to four decimal places.)

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

+3
Answers (1)
  1. 19 September, 21:21
    0
    a) Bi [P (X >=15) ] ≈ 0.9944

    b) Bi [P (X >=30) ] ≈ 0.3182

    c) Bi [P (25=< X = < 35) ] ≈ 0.6623

    d) Bi [P (X >40) ] ≈ 0.0046

    Step-by-step explanation:

    Given:

    - Total sample size n = 745

    - The probability of success p = 0.037

    - The probability of failure q = 0.963

    Find:

    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

    Solution:

    - The condition for normal approximation to binomial distribution:

    n*p = 745*0.037 = 27.565 > 5

    n*q = 745*0.963 = 717.435 > 5

    Normal Approximation is valid.

    a) P (X > = 15) ?

    - Apply continuity correction for normal approximation:

    Bi [P (X >=15) ] = N [ P (X > = 14.5) ]

    - Then the parameters u mean and σ standard deviation for normal distribution are:

    u = n*p = 27.565

    σ = sqrt (n*p*q) = sqrt (745*0.037*0.963) = 5.1522

    - The random variable has approximated normal distribution as follows:

    X~N (27.565, 5.1522^2)

    - Now compute the Z - value for the corrected limit:

    N [ P (X > = 14.5) ] = P (Z > = (14.5 - 27.565) / 5.1522)

    N [ P (X > = 14.5) ] = P (Z > = - 2.5358)

    - Now use the Z-score table to evaluate the probability:

    P (Z > = - 2.5358) = 0.9944

    N [ P (X > = 14.5) ] = P (Z > = - 2.5358) = 0.9944

    Hence,

    Bi [P (X >=15) ] ≈ 0.9944

    b) P (X > = 30) ?

    - Apply continuity correction for normal approximation:

    Bi [P (X >=30) ] = N [ P (X > = 29.5) ]

    - Now compute the Z - value for the corrected limit:

    N [ P (X > = 29.5) ] = P (Z > = (29.5 - 27.565) / 5.1522)

    N [ P (X > = 29.5) ] = P (Z > = 0.37556)

    - Now use the Z-score table to evaluate the probability:

    P (Z > = 0.37556) = 0.3182

    N [ P (X > = 29.5) ] = P (Z > = 0.37556) = 0.3182

    Hence,

    Bi [P (X >=30) ] ≈ 0.3182

    c) P (25=< X = < 35) ?

    - Apply continuity correction for normal approximation:

    Bi [P (25=< X = < 35) ] = N [ P (24.5=< X = < 35.5) ]

    - Now compute the Z - value for the corrected limit:

    N [ P (24.5=< X = < 35.5) ] = P ((24.5 - 27.565) / 5.1522 =
    N [ P (24.5=< X = < 25.5) ] = P (-0.59489 =
    - Now use the Z-score table to evaluate the probability:

    P (-0.59489 =
    N [ P (24.5=< X = < 35.5) ] = P (-0.59489 =
    Hence,

    Bi [P (25=< X = < 35) ] ≈ 0.6623

    d) P (X > 40) ?

    - Apply continuity correction for normal approximation:

    Bi [P (X >40) ] = N [ P (X > 41) ]

    - Now compute the Z - value for the corrected limit:

    N [ P (X > 41) ] = P (Z > (41 - 27.565) / 5.1522)

    N [ P (X > 41) ] = P (Z > 2.60762)

    - Now use the Z-score table to evaluate the probability:

    P (Z > 2.60762) = 0.0046

    N [ P (X > 41) ] = P (Z > 2.60762) = 0.0046

    Hence,

    Bi [P (X >40) ] ≈ 0.0046
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers