Question

What should I buy? A study conducted by a research group in a recent year reported that

57%

of cell phone owners used their phones inside a store for guidance on purchasing decisions. A sample of

15

cell phone owners is studied. Round the answers to at least four decimal places.

Part 1 of 4

(a) What is the probability that five or more of them used their phones for guidance on purchasing decisions?

The probability that five or more of them used their phones for guidance on purchasing decisions is .

Part 2 of 4

(b) What is the probability that fewer than eight of them used their phones for guidance on purchasing decisions?

The probability that fewer than eight of them used their phones for guidance on purchasing decisions is .

Part 3 of 4

(c) What is the probability that exactly nine of them used their phones for guidance on purchasing decisions?

The probability that exactly nine of them used their phones for guidance on purchasing decisions is .

Part 4 of 4

(d) Would it be unusual if more than

11

of them used their phones for guidance on purchasing decisions?

It ▼(Choose one) be unusual if more than

11

of them used their phones for guidance on purchasing decisions since the probability is .

Answer 1

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 = 0.57

N = number of trials = 15

R = desired success

A)

P(5 or more) = 1 - (p(0) + p(1) + p(2) + p(3) + p(4)) = 0.98259022376

B)

P(fewer than 8) = P(0) + .... p(7) = 0.28976586616

C)

P(9) = 15c9*(0.57^9)*(1-0.57)^15-9 = 0.20095023404

D)

P(more than 11) = P(12) + P(13) + P(14) + P(15) = 0.0582509488

Since the probability is greater than 0.05.

It is not unusal.