Respuesta :
Answer:
(a) The mean, variance and standard deviation of random variable X are 20, 4 and 2 respectively.
(b) The probability that X is within one standard deviation of the mean is 0.7926.
(c) The probability that X is more than two standard deviations from the mean is 0.0038.
Step-by-step explanation:
The random variable X follows a Binomial distribution with parameter n = 25 and p = 0.80.
The probability mass function of X is:
[tex]P(X=x)={25\choose x}0.80^{x}(1-0.80)^{25-x};\ x=0,1,2,3...[/tex]
The mean, variance and standard deviation of a Binomial distribution is given by:
[tex]\mu=np\\\sigma^{2}=np(1-p)\\\sigma=\sqrt{np(1-p)}[/tex]
(a)
Compute the mean, variance and standard deviation of random variable X as follows:
[tex]\mu=np=25\times0.80=20\\\sigma^{2}=np(1-p)=25\times0.80\times(1-0.80)=4\\\sigma=\sqrt{np(1-p)}=\sqrt{4}=2[/tex]
Thus, the mean, variance and standard deviation of random variable X are 20, 4 and 2 respectively.
(b)
Compute the probability that X is within one standard deviation of the mean as follows:
P (μ - σ ≤ X ≤ μ + σ) = P (20 - 2 ≤ X ≤ 20 + 2)
= P (18 ≤ X ≤ 22)
= P (X ≤ 22) - P (X ≤ 18)
= P (X = 18) + P (X = 19) + P (X = 20) + P (X = 21) + P (X = 22)
[tex]={25\choose 18}0.80^{18}(1-0.80)^{25-18}+{25\choose 19}0.80^{19}(1-0.80)^{25-19}\\+{25\choose 20}0.80^{20}(1-0.80)^{25-20}+{25\choose 21}0.80^{21}(1-0.80)^{25-21}\\+{25\choose 22}0.80^{22}(1-0.80)^{25-22}\\=0.1108+0.1633+0.1960+0.1867+0.1358\\=0.7926[/tex]
Thus, the probability that X is within one standard deviation of the mean is 0.7926.
(c)
Compute the probability that X is more than two standard deviations from the mean as follows:
P (X > μ + 2σ) = P (X > 20 + (2×2))
= P (X > 24)
= P (X = 25)
[tex]={25\choose 25}0.80^{25}(1-0.80)^{25-25}\\=1\times 0.00378\times1\\=0.0038[/tex]
Thus, the probability that X is more than two standard deviations from the mean is 0.0038.
