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tensa zangetsu [6.8K]
2 years ago
12

OK people stop putting your questions in the awnser box because you are just using poibts that can go to other people who actual

ly wanted to answer
Mathematics
1 answer:
Lyrx [107]2 years ago
5 0

Answer:

yes ma'am

Step-by-step explanation:

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In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist
Marrrta [24]

Answer:

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 =<Z =< (35.5 - 27.565) / 5.1522 )

                N [ P ( 24.5=< X =< 25.5 ) ] = P ( -0.59489 =<Z =< 1.54011 )

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

                P ( -0.59489 =<Z =< 1.54011 ) = 0.6623

               N [ P ( 24.5=< X =< 35.5 ) ]= P ( -0.59489 =<Z =< 1.54011 ) = 0.6623

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  

4 0
3 years ago
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