Respuesta :
Answer:
a ) 0.1403604645 and 0.1368
b) 0.3464961 and 0.3485
c) 0.802671982 and 0.8018
Step-by-step explanation:
Y~ B (15,0.45)
Y~ N (15*0.45, 15*0.45*0.55) = Y~ N (6.75, 3.7125)
a) P(Y=5) = 15C5 (0.45)^5 * (0.55)^10 = 0.1403604645
For normal approximation
P(Y = 5 ) = P ( 4.5 < Y < 5.5 ) ......... continuity correction
Hence,
[tex]P ( 4.5 < Y < 5.5 ) = P ( \frac{4.5 - 6.75}{\sqrt{3.7125} } < Z < \frac{5.5 - 6.75}{\sqrt{3.7125} } ) = P ( -1.16775 < Z < -0.64875 )[/tex]
The probability P ( 4.5 < Y < 5.5 ) = 0.1368
b) P(Y>7) = 15C8 (0.45)^ 8 (0.55)^7 + 15C9 (0.45)^9 * (0.55)^6 + 15C10 (0.45)^10 * (0.55)^5 + 15C11 (0.45)^11 * (0.55)^4 + 15C12 (0.45)^12 * (0.55)^3 + 15C13 (0.45)^13 * (0.55)^2 + 15C14 (0.45)^14 * (0.55) + (0.45)^15
= 0.3464961
For normal approximation
P(Y > 7 ) = P (Y > 7.5 ) ......... continuity correction
Hence,
[tex]P (Y > 7.5) = P (Z > \frac{7.5-6.75}{\sqrt{3.7125} } ) = P (Z > 0.389249)\\[/tex]
The probability P ( Y>7.5 ) = 0.3485
c) P (4 < Y < 10) = 15C5 (0.45)^5 (0.55)^10 + 15C6 (0.45)^ 6 (0.55)^9 + 15C7 (0.45)^7 (0.55)^8 + 15C8 (0.45)^ 8 (0.55)^7 + 15C9 (0.45)^9 * (0.55)^6
= 0.802671982
For normal approximation
P( 4 < Y < 10 ) = P (4.5< Y < 9.5 ) ......... continuity correction
Hence,
[tex]P ( 4.5 < Y < 9.5 ) = P ( \frac{4.5 - 6.75}{\sqrt{3.7125} } < Z < \frac{9.5 - 6.75}{\sqrt{3.7125} } ) = P ( -1.167748416 < Z < 1.427248064 )[/tex]
The probability P (4.5< Y < 9.5 ) = 0.8018
