Question

According to a recent survey, 73% of all customers will return to the same grocery store. Suppose 14 customers are selected at random, what is the probability that: (Round probabilities to 4 decimal places and the expected customers to 2 decimal places)

Answer 1

This question is incomplete, the complete question is;

According to a recent survey, 73% of all customers will return to the same grocery store. Suppose 14 customers are selected at random, what is the probability that: (Round probabilities to 4 decimal places and the expected customers to 2 decimal places)

a) All 14 will return?

b) How many customers would be expected to return to the store ?

Answer:

a) the probability that All 14 will return is 0.0122

b) The number of customers that would be expected to return to the store is 13.27

Step-by-step explanation:

Given the data in the question;

p = 73% = 0.73

sample size n = 14

a)  the probability that All 14 will return;

P( x = 14 ) = ⁿCₓ × Pˣ × $( 1 - P)^{n-x$

we substitute

P( x = 14 ) = ¹⁴C₁₄ × (0.73)¹⁴ × $( 1 - 0.73)^{(14-14)$

P( x = 14 ) = [ 14! / ( 14! × ( 14 - 14)!) ] ×  (0.73)¹⁴ × $( 0.27)^{0$

P( x = 14 ) = 1 × 0.0122045 × $( 0.27)^{0$

P( x = 14 ) = 1 × 0.0122045 × 1

P( x = 14 ) = 0.0122045 ≈ 0.0122 { 4 decimal place }

Therefore, the probability that All 14 will return is 0.0122

b) Expected Customers;

E(x) = n × p

we substitute

E(x) = 14 × 0.73

E(x) =  13.27  { 2 decimal place }

Therefore, The number of customers that would be expected to return to the store is 13.27