Respuesta :
Given:
p = 47% = 0.47, the probability that a man considers himself a professional baseball fan.
q = 1 - p = 0.53, the probability that a man does not consider himself a professional baseball fan.
n = 10, the number of men surveyed.
(a) Calculate the probability that exactly 5 of 10 men surveyed consider themselves as professional baseball fans.
P(5 of 10) = ₁₀C₅ p⁵q⁵ = 252*0.47⁵*0.53⁵ = 0.242
Answer: 0.242 or 24.2%
(b) Calculate the probability of at least 6 out of 10.
P(at least 6 of 10)
= ₁₀C₆ p⁶q⁴ + ₁₀C₇ p⁷q³ + ₁₀C₈ p⁸q² + ₁₀C₉ p⁹q + ₁₀C₁₀ p¹⁰ q⁰
= 0.1786 + 0.0905 + 0.0301 + 0.0059 + 0.0
= 0.3057
Answer: 0.3057 or 30.6%
(c) Calculate the probability of less than 4 of 10.
P(less than 4 of 10)
= ₁₀C₁ pq⁹ + ₁₀C₂ p²q⁸ + ₁₀C₃ p³q⁷
= 0.0155 + 0.0619 + 0.1464
= 0.2238
Answer: 0.2238 or 22.4%
p = 47% = 0.47, the probability that a man considers himself a professional baseball fan.
q = 1 - p = 0.53, the probability that a man does not consider himself a professional baseball fan.
n = 10, the number of men surveyed.
(a) Calculate the probability that exactly 5 of 10 men surveyed consider themselves as professional baseball fans.
P(5 of 10) = ₁₀C₅ p⁵q⁵ = 252*0.47⁵*0.53⁵ = 0.242
Answer: 0.242 or 24.2%
(b) Calculate the probability of at least 6 out of 10.
P(at least 6 of 10)
= ₁₀C₆ p⁶q⁴ + ₁₀C₇ p⁷q³ + ₁₀C₈ p⁸q² + ₁₀C₉ p⁹q + ₁₀C₁₀ p¹⁰ q⁰
= 0.1786 + 0.0905 + 0.0301 + 0.0059 + 0.0
= 0.3057
Answer: 0.3057 or 30.6%
(c) Calculate the probability of less than 4 of 10.
P(less than 4 of 10)
= ₁₀C₁ pq⁹ + ₁₀C₂ p²q⁸ + ₁₀C₃ p³q⁷
= 0.0155 + 0.0619 + 0.1464
= 0.2238
Answer: 0.2238 or 22.4%
The situation represents a binomial probabilty with the probability of success (p) = 47% or 0.47 and the number of trials (n) = 10.
The probability of a binomial distribution is given by:
[tex]P(X)=\left(^n_x)p^x(1-p)^{n-x}[/tex]
Part A:
The probability that the number who consider themselves baseball fans is exactly five is given by:
[tex]P(5)=\left(^{10}_{\,5}\right)(0.47)^5(1-0.47)^{10-5} \\ \\ =252(0.022935)(0.041820)=\bold{0.2417}[/tex]
Part B:
The probability that the number who consider themselves baseball fans is at least six is given by:
[tex]P(X\geq6)=P(6)+P(7)+P(8)+P(9)+p(10) \\ \\ =\left(^{10}_{\,6}\right)(0.47)^6(1-0.47)^{10-6}+\left(^{10}_{\,7}\right)(0.47)^7(1-0.47)^{10-7} \\ +\left(^{10}_{\,8}\right)(0.47)^8(1-0.47)^{10-8}+\left(^{10}_{\,9}\right)(0.47)^9(1-0.47)^{10-9} \\ +\left(^{10}_{10}\right)(0.47)^{10}(1-0.47)^{10-10} \\ \\ =210(0.010779)(0.078905)+120(0.005066)(0.148877) \\ +45(0.002381)(0.2809)+10(0.001119)(0.53)+1(0.000526)(1) \\ \\ =0.1786+0.0905+0.0301+0.0059+0.0005=\bold{0.3056}[/tex]
This can be approximated using normal distribution as I will illustrate in part c.
Part C:
The probability that the number who consider themselves baseball fans is less than four is given by:
[tex]P(X\ \textless \ 4)=P(0)+P(1)+P(2)+P(3)[/tex]
We can approximate this using normal distribution by subtracting 0.5 from the least value and adding 0.5 to the greatest value.
This gives [tex]P(0-0.5\ \textless \ X\ \textless \ 3+0.5)=P(-0.5\ \textless \ X\ \textless \ 3.5)[/tex]
The mean of a binomial distribution is given by [tex]\mu=np[/tex], while the standard deviation is given by [tex]\sigma= \sqrt{np(1-p)} [/tex].
Thus,
[tex]\mu=10(0.47)=4.7 \ and \ \sigma=\sqrt{10(0.47)(0.53)}=1.578[/tex]
The probability of a nomal distribution between two values (a, b) is given by:
[tex]P(a\ \textless \ X\ \textless \ b)=P\left( \frac{b-\mu}{\sigma} \right)-P\left( \frac{a-\mu}{\sigma} \right)[/tex]
Thus,
[tex]P(-0.5\ \textless \ X\ \textless \ 3.5)=P\left( \frac{3.5-4.7}{1.578} \right)-P\left( \frac{-0.5-4.7}{1.578} \right) \\ \\ =P(-0.7605)-P(-3.295)=0.2235-0.0005=\bold{0.223}[/tex]
The probability of a binomial distribution is given by:
[tex]P(X)=\left(^n_x)p^x(1-p)^{n-x}[/tex]
Part A:
The probability that the number who consider themselves baseball fans is exactly five is given by:
[tex]P(5)=\left(^{10}_{\,5}\right)(0.47)^5(1-0.47)^{10-5} \\ \\ =252(0.022935)(0.041820)=\bold{0.2417}[/tex]
Part B:
The probability that the number who consider themselves baseball fans is at least six is given by:
[tex]P(X\geq6)=P(6)+P(7)+P(8)+P(9)+p(10) \\ \\ =\left(^{10}_{\,6}\right)(0.47)^6(1-0.47)^{10-6}+\left(^{10}_{\,7}\right)(0.47)^7(1-0.47)^{10-7} \\ +\left(^{10}_{\,8}\right)(0.47)^8(1-0.47)^{10-8}+\left(^{10}_{\,9}\right)(0.47)^9(1-0.47)^{10-9} \\ +\left(^{10}_{10}\right)(0.47)^{10}(1-0.47)^{10-10} \\ \\ =210(0.010779)(0.078905)+120(0.005066)(0.148877) \\ +45(0.002381)(0.2809)+10(0.001119)(0.53)+1(0.000526)(1) \\ \\ =0.1786+0.0905+0.0301+0.0059+0.0005=\bold{0.3056}[/tex]
This can be approximated using normal distribution as I will illustrate in part c.
Part C:
The probability that the number who consider themselves baseball fans is less than four is given by:
[tex]P(X\ \textless \ 4)=P(0)+P(1)+P(2)+P(3)[/tex]
We can approximate this using normal distribution by subtracting 0.5 from the least value and adding 0.5 to the greatest value.
This gives [tex]P(0-0.5\ \textless \ X\ \textless \ 3+0.5)=P(-0.5\ \textless \ X\ \textless \ 3.5)[/tex]
The mean of a binomial distribution is given by [tex]\mu=np[/tex], while the standard deviation is given by [tex]\sigma= \sqrt{np(1-p)} [/tex].
Thus,
[tex]\mu=10(0.47)=4.7 \ and \ \sigma=\sqrt{10(0.47)(0.53)}=1.578[/tex]
The probability of a nomal distribution between two values (a, b) is given by:
[tex]P(a\ \textless \ X\ \textless \ b)=P\left( \frac{b-\mu}{\sigma} \right)-P\left( \frac{a-\mu}{\sigma} \right)[/tex]
Thus,
[tex]P(-0.5\ \textless \ X\ \textless \ 3.5)=P\left( \frac{3.5-4.7}{1.578} \right)-P\left( \frac{-0.5-4.7}{1.578} \right) \\ \\ =P(-0.7605)-P(-3.295)=0.2235-0.0005=\bold{0.223}[/tex]
