Question

Although cities encourage carpooling to reduce traffic congestion, most vehicles carry only one person. For example, 79% of vehicles on the roads are occupied by just the driver. (Round your answers to four decimal places.)

(a) If you choose 10 vehicles at random, what is the probability that more than half (that is, 6 or more) carry just one person?


(b) If you choose 108 vehicles at random, what is the probability that more than half (that is, 55 or more) carry just one person? (Use the normal approximation.)

Answer 1

A)

Answer)

As there are fixed number of trials and probability of each and every trial is same and independent of each other

Here we need to use the binomial formula

P(r) = ncr*(p^r)*(1-p)^n-r

Ncr = n!/(r!*(n-r)!)

N! = N*n-1*n-2*n-3*n-4*n-5........till 1

For example 5! = 5*4*3*2*1

Special case is 0! = 1

P = probability of single trial = 0.79

N = number of trials = 10

R = desired success = 6 or more

P(6 or more) = p(6) + p(7) + p(8) + p(9) + p(10)

= 0.96013760908

= 0.9601

B)

N = 108

P = 0.79

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 85.32

N*(1-p) = 22.68

Both the conditions are met so we can use standard normal z table to estimate the probability

Z = (x - mean)/s.d

Mean = n*p = 85.32

S.d = √{n*p*(1-p)} = 4.23287136586

P(x>=55)

By continuity correction

P(x>54.5)

Z = (54.5 - 85.32)/4.23287136586

Z = -7.28

From z table, P(z>-7.28) = 0.9999 ~ 1