On average, there are two customers at the supermarket checkout every minute, and the probability of no customers checking out in one minute.(rounded to four decimal places).
Question
On average, there are two customers at the supermarket checkout every minute, and the probability of no customers checking out in one minute.(rounded to four decimal places).
Solution
This is a Poisson distribution problem. The Poisson distribution gives the probability of a given number of events (in this case, customers checking out) happening in a fixed interval of time (in this case, one minute) when these events occur with a known average rate (in this case, 2 customers per minute) and independently of the time since the last event.
The formula for the Poisson probability is:
P(X=k) = (λ^k * e^-λ) / k!
where:
- P(X=k) is the probability of k events in the interval
- λ is the average rate of events per interval
- e is the base of the natural logarithm, approximately equal to 2.71828
- k! is the factorial of k
In this case, we want to find the probability of no customers (k=0) checking out in one minute, with an average rate of 2 customers per minute (λ=2).
So, we plug these values into the formula:
P(X=0) = (2^0 * e^-2) / 0! = (1 * e^-2) / 1 = e^-2 = 0.1353 (rounded to four decimal places)
So, the probability of no customers checking out in one minute is approximately 0.1353.
Similar Questions
The manager of a grocery store has taken a random sample of 100 customers to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes. She observes that the average length of time it took the customers in the sample to check out was 3.1 minutes. The population standard deviation is known to be 0.5 minutes.
The waiting time to check out of a supermarket has had a population mean of 11.02 minutes.Recently,in an effort to reduce the waiting time,the supermarket has experimented with a system in which there is a single waiting line with multiple checkout servers.A sample of 50 customers was selected,and their mean waiting time to check out was 9.79 minutes with a sample standard deviation of 5.6 minutes.Complete parts(a)through(d). Click here to view page 1 of the critical values for the t distribution. Click here to view page 2 of the critical values for the t Distribution. a.At the 0.01 level of significance,using the critical value approach to hypothesis testing,is there evidence that the population mean waiting time to check out is less than 11.02 minutes? What are the null and alternative hypotheses for this test?
In a bank an average of 5 out of every 10 customers ask for opening a new account. A random sample of 20 customers is selected. Calculate the probability that exact 7 customers ask for opening a new account. Round your answer four decimal places.
The amount of time that customers wait in line during peak hours at one bank is normally distributed with a mean of 12 minutes and a standard deviation of 2 minutes. What is the probability that a customer will wait under 14 minutes?
Customers arrive at a drive-up window of a fast-food restaurant at a rate of 3 per minute during lunch hour. a.What is the probability that the next customer will arrive within 2 minutes? b.What is the probability that the next customer will arrive within 4 minutes? c.During the dinner time period,the arrival rate is 1 per minute.What are your answers to a and b for this period? Round to four decimal places as needed.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.