A survey across banks in Delhi showed the mean time taken to serve customers was 6.4...
A survey across banks in Delhi showed the mean time taken to serve customers was 6.4 minutes while the standard deviation was 1.5 minutes. It is not known whether the data distribution is symmetrical. a) What is the probability of a customer being served between 2.8 and 10 minutes? b) What is the probability of a customer being served at the expected value of the distribution? Show the sketches for your answers.
Homework Answers
It is given that the survey across banks in Delhi showed the mean time taken to serve customers was 6.4 minutes while the
standard deviation was 1.5 minutes.
Also here we don't known whether the data distribution is symmetrical.
So we need to use Chebyshev's inequality to find the probability.
a) Here we want to find the probability of a customer being served between 2.8 and 10 minutes
Here 2.8 - 6.4 = -3.6
and 10 - 6.4 = 3.6
So both the values 2.8 and 10 are same far from the mean of the distribution.
So the probability between 2.8 and 10 minutes is at least
Let's find the value of k
Therefore required probability is as follows
b) What is the probability of a customer being served at the expected value of the distribution? Show the sketches for your answers.
The expected value of the distribution is same as mean = 6.4
So P(X = 6.4) = 0
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