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
