Question

Part Two. Can you help me with the binomial probability distribution? Please see instructions below.

Please show all steps to understand better. Thank you in advance.

5. Use the binomial distribution table to find the binomial probabilities for these cases: a) n=10, p=1, k= 3 b) n=14, p = 6, k = 7 c) n =25, p=5, k = 14 6. Consider a binomial experiment with n = 20 and p=0.70. Use the binomial formula to calculate: a) P(x = 15) b) mean c) variance d) standard deviation 7. 35% of YALE University students use their own car to get to college. What is the probability that, in a sample of 15 students, four (4) students use their own car? (Use the binomial formula to solve the exercise). 8. 40% of Americans traveling by car search for gas stations and grocery stores that are near or visible from the road. Suppose a random sample of n=25 Americans who travel by car are asked how they determine where to stop to buy food and get gas. Let x denote the number of the responding sample looking for gas stations and grocery stores that are near or visible from the highway. (Use the binomial formula to solve the exercise). Answers: a) What is the mean if x = 10? b) What is the variance of x? c) Calculate the interval u – 20. d) What values of the binomial random variable x fall in this interval?

Answer 1

Solution : The binomial probability distribution is distribution of occurance of number k out of n observation.

suppose a coin is tossed 10 times and probability of getting head is 0.5 then we find probability for k times we get head out of n=10 times .

5. a) given n=10, p=0.1 and k=3

from table, we look for n=10 in row and p=0.1 and k=3 ,

we get,

P[k = 3]=0.05739

b) given n=14, p=0.6 and k=7

from table,  

P[k = 7]= 0.15740

c) given n=25, p=0.5 and k=14

from table,  

P[k = 14] = 0.132840

6. Given that n= 20 and p= 0.7

a) we have to find the probability for x taking value 15

we know formula,

P(x=15) = nCx p^x q^n-x

=( 20 C 15) * (0.7) ¹⁵ (0.3) ^20-15

= 0.178863

Thus, the probability of X=15 is 0.178863

b) here p=0.7 and q= 1-p = 1-0.7= 0.3

q=0.3

we know mean = np

= 20*0.7

=14

c) to find varience we have

var= npq

= 20*0.7*0.3

=4.2

d) The squre root of varience is sd

thus sd = 2.0493

7) Given that the 0.35% student use their car

thus p= 0.35 and the sample size n is 15

now we have to find probability that , out of those 15 students 4 student use their car

x=4

P[x=4] = nCx p^x q^n-x

=( 15 C 4) * (0.35) ⁴ (0.65) ^15-4

= 0.179246

Thus, probability that , out of those 15 students 4 student use their car is 0.1792

8) Given that p=0.4 and n= 25

let X denote the number of responding sample that are looking for the gas and food nearby the highway

sample size is 25 and p=0.4

a) The mean is

mean = np = 25*0.4

= 10

The mean is not depend on the value of X

b) the var of X is ,

var= npq

= 25 *0.4*0.6

=6

and sd is 2.449

c) Now the CI is

lb = $\mu - 2*\sigma$

= 10 - 2* 2.449

= 5.1010

and ub = $\mu - 2*\sigma$

= 10 + 2* 2.449

= 14.898

Thus the CI is ( 5.1010, 14.898 )

d) The given interval is 2 sigma interval and we know that 95.65% observations do lie in the 2 sigma interval.