Binomial Distribution & Poisson Distribution

[satyavir]

Dear All,

I got stuck in below 2 probability questions & wanted to check if anybody from this forum could help me out. I know that the first one could be solved using Poisson Distribution Formula & the 2nd one by Binomial. Some workings mentioned below:

1. The number of customers who enter a ‘German’ supermarket-Gandhinagar each hour is ‘normally’ distributed with a mean of 600 and a standard deviation of 200. The supermarket is open 16 hours per day. What is the probability that the total number of customers who enter the supermarket in one day is greater than 10,000?

My Working:

Lambda = 600 customers every 1 hour
x = 10000+ customers in 16 hours

Now as per the rule, we can't change the x interval, hence if we change the lambda value to reconcile it as:

Lambda = 9600 customers in 16 hours
x = 10000+ customers in 16 hours

Formula: P (X =x) = (e– λ λx)/x! ??


2. Shree Ganga Taploo University bookstore claims that 50% of its customers are satisfied with the service and prices. If this claim is true, what is the probability that in a random sample of 600 customers less than 45% are satisfied with services and price?

My Working: p = .50, q = .50 (1-p), n = 600, x<45

Now, if I try to put this as per the below Binomial formula, the value of n = 600 seems huge plus the question says less than 45% which as per my understanding means finding value of x for 45, 44, 43,42..........0) which will be a cumbersome task

 

[BigBeachBanana]

Dear All,

I got stuck in below 2 probability questions & wanted to check if anybody from this forum could help me out. I know that the first one could be solved using Poisson Distribution Formula & the 2nd one by Binomial. Some workings mentioned below:

1. The number of customers who enter a ‘German’ supermarket-Gandhinagar each hour is ‘normally’ distributed with a mean of 600 and a standard deviation of 200. The supermarket is open 16 hours per day. What is the probability that the total number of customers who enter the supermarket in one day is greater than 10,000?

My Working:

Lambda = 600 customers every 1 hour
x = 10000+ customers in 16 hours

Now as per the rule, we can't change the x interval, hence if we change the lambda value to reconcile it as:

Lambda = 9600 customers in 16 hours
x = 10000+ customers in 16 hours

Formula: P (X =x) = (e– λ λx)/x! ??


2. Shree Ganga Taploo University bookstore claims that 50% of its customers are satisfied with the service and prices. If this claim is true, what is the probability that in a random sample of 600 customers less than 45% are satisfied with services and price?

My Working: p = .50, q = .50 (1-p), n = 600, x<45

Now, if I try to put this as per the below Binomial formula, the value of n = 600 seems huge plus the question says less than 45% which as per my understanding means finding value of x for 45, 44, 43,42..........0) which will be a cumbersome task

#1) They told you that it's normally distributed, so try again with the normal distribution. Also, notice the mean and variance are not equal, so it can't be Poisson.

#2) For a large sample size n, you can approximate the binomial with the normal distribution with continuity correction. Sounds familiar?

PS: Next time, please post one question per thread.