Answer:
a.
[tex]P(X=3)=0.2903\\\\P(X \geq 3)=0.5801\\\\P(X\leq 3)0.7102[/tex]
b.
[tex]P(2\leq x\leq 4 )=0.7451[/tex]
c. mean=2.8
d . standard deviation=1.2961
Step-by-step explanation:
We determine that the accident rates follow a binomial distribution. The rate of success p=0.4 and sample n=7:
[tex]P(x)={n\choose x}p^x(1-p)^{n-x}[/tex]
#the probability of exactly three;
[tex]P(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X=3)={7\choose 3}0.4^3(0.6)^{4}\\\\=0.2903[/tex]
#At least(more than 2)
[tex]P(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 3)=1-P(X\leq 2)\\\\=1-{7\choose 0}0.4^0(0.6)^{7}-{7\choose 1}0.4^1(0.6)^{6}-{7\choose 2}0.4^2(0.6)^{5}\\\\=1-0.0280-0.1306-0.2613\\\\=0.5801[/tex]
#At most 3;
[tex]P(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X \leq 3)={7\choose 0}0.4^0(0.6)^{7}+{7\choose 1}0.4^1(0.6)^{6}+{7\choose 2}0.4^2(0.6)^{5}+{7\choose 3}0.4^3(0.6)^{4}\\\\\\\=0.0280+0.1306+0.2613+0.2903\\\\=0.7102[/tex]
b. Between 2 and 4:
Using the binomial expression, this probability is calculated as:
[tex]P(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(2\leq x\leq 4 )={7\choose 2}0.4^2(0.6)^{5}+{7\choose 3}0.4^3(0.6)^{4}+{7\choose 4}0.4^4(0.6)^{3}\\\\\\\\\=0.2613+0.2903+0.1935\\\\=0.7451[/tex]
Hence,the probability of between 2 and four is 0.7451
c. From a above, we have the values of n=7 and p=0.4.
-We substitute this values in the formula below to calculate the mean:
-The mean of a binomial distribution is calculated as the product of the probability of success by the sample size, mean=np:
[tex]\mu=np, n=7, p=0.4\\\\\mu=7\times 0.4\\\\=2.8[/tex]
Hence, the standard deviation of the sample is 2.8
d. From a above, we have the values of n=7 and p=0.4
--We substitute this values in the formula below to calculate the standard deviation
-The standard deviation a binomial distribution is given as:
[tex]\sigma={\sqrt {np(1-p)}\\\\=\sqrt{7\times 0.4\times 0.6}\\\\=1.2961[/tex]
Hence, the standard deviation of the sample is 1.2961
