Answer:
a) 0.2816
b) 0.019698
c)0.755974
d)E(X)=2.5
e) σ =1.369
Step-by-step explanation:
a. This is a binomial probability distribution with n=10 and p=25%=0.25
The binomial probability function is denoted as:
[tex]P(X=x)={n\choose x}p^x(1-p)^{n-x}[/tex]
#Probability that Exactly 5 adults own tablets computers:
[tex]P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X=2)={10\choose 2}0.25^2(1-0.25)^8\\\\=0.2816[/tex]
Hence, the Probability that Exactly 5 adults own tablets computers is 0.2816
b.Between 6 and 8 adults (inclusive) own tablet computers:
-We use our function from a above to find the summation of probabilities for x=6,7,8:
[tex]P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(6\leq X\leq 8)=P(X=6)+P(X=7)+P(X=8)\\\\={10\choose6 }0.25^6(1-0.25)^4+{10\choose7}0.25^7(1-0.25)^3+{10\choose8 }0.25^8(1-0.25)^2\\\\=0.016222+0.003090+0.000386\\\\=0.019698[/tex]
Hence, the probability that between 6 and 8 adults (inclusive) own tablet computers is 0.019698
c. At least 2 adults own tablet computers.
We use our function to find the probabilities for x=2,3,4,5,6,7,8,9,10 and sum them:
[tex]P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq2 )=1-(P(X=0)+P(X=1))\\\\=1-[{10\choose0 }0.25^0(1-0.25)^{10}+{10\choose1}0.25^1(1-0.25)^9]\\\\=1-[0.056314+0.187712]\\\\=0.755974[/tex]
Hence, the probability that At least 2 adults own tablet computers is 0.755974
d. The expected value in a binomial distribution is the product of probability of success by the sample size, n;
[tex]E(X)=\mu=np, n=10 p=0.25\\\\=10\times 0.25\\\\=2.5[/tex]
Hence, the expected value of the function is 2.5
e. The standard deviation of a binomial distribution function is calculated as:
[tex]\sigma=\sqrt{np ( 1 - p) }, n=10, p=0.25\\\\=\sqrt{10\times 0.25\times(1-0.25)}\\\\=1.369[/tex]
Hence, the standard deviation of the function is 1.369
