Suppose that 75% of all trucks undergoing a brake inspection at a certain inspection facility pass the inspection. Consider groups of 15 trucks and let X be the number of trucks in a group that have passed the nspection Part a: Verify that this is a binomial distribution Part b: For what proportion of groups will between 8 and 10 trucks (inclusive) pass the inspection? Part c: For what proportion of groups will exactly 3 trucks fail the inspection? Part d: Find the mean and the standard deviation of the random variable X

b) 0.2961

c) 0.2251

d) Mean = 11.25, Standard deviation =  1.667  

Step-by-step explanation:

We are given the following information:

We treat trucks undergoing a brake inspection passin as a success.

P( trucks undergoing a brake inspection passes the test) = 75% = 0.75

a) Conditions for binomial probability distribution

Then the number of trucks undergoing a brake inspection follows a binomial distribution, where

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 15

b) P(proportion of groups will between 8 and 10 trucks pass the inspection)

We have to evaluate:

c) P( exactly 3 trucks fail the inspection)

p = 0.25

d) Mean and standard deviation

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2

the solution is the point

step-by-step explanation:

we have

> equation a

> equation b

adds equation a and equation b

substitute the value of x in equation b to find the value of y

the solution is the point