Based on past surveys over several years, a small business believes that 75% of their customers are satisfied with their services. Due to the recent conditions, changes in policy and additional charges, the business decides to select a random sample of 16 recent customers and ask them if they are satisfied with their services. If the number of satisfied customers in the sample is 11 or fewer, the business will conclude that this is an indication of a need for improvement in customer service and they will try to determine ways to provide it. What is the probability that the results of the sample will indicate a need to improve customer service?

The probability that the results of the sample will indicate a need to improve customer service is 0.3698138

Step-by-step explanation

This requires the application of the binomial distribution which which models a discrete random variable representing the number of successes in the n trials. 

Interpretation :  In this case the sample consists of 16 customers. There are two possible outcomes for each customer i.e he/she may either be satisfied  or not satisfied with the services.  The probability that a customer is satisfied is 0.75 and the probability that a customer is not satisfied is 1 - 0.75 = 0.25. 

Let X be a random variable representing the number of customers in the sample  who are satisfied. Then X follows a binomial distribution with parameters n=16 and p=0.75 

The task to compute probability that the number of satisfied customers in the sample is 11 or fewer

                         P(X≤11)=P(X=0)+P(X=1)+...+P(X=11) 

                         =(016?)0.750(0.25)16+(116?)0.751(0.25)15+(216?)0.752(0.25)14+...+(1116?)0.7511(0.25)5

                        = 2.328306×10−10+1.117587×10−8+2.514571×10−7+3.520399×10−6+3.432389×10−5+2.471320×10−4+1.359226×10−3+5.825255+1.966024×10−2+5.242730×10−2+1.100973×10−1+1.801593×10−1=0.3698138

Using R, this can be computed  easily as follows 

> pbinom(11,16,0.75)
[1] 0.3698138