Question

A supermarket has determined that daily demand for one-liter milk bottles has an approximate bell shaped distribution, with a mean of 55 bottles and a variance of 36

a) How often can we expect between 43 and 67 bottles to be sold in a day? (Give a percentage.)

b) What percentage of the time will the number of bottles sold more than 61 bottles

c) If the supermarket begins each morning with a supply of 73 bottles how often will demand exceed the supply? (Give a percentage.)

Answer 1

Solution:
Given in the question
A supermarket has determined that daily demand for one-liter milk bottles has an approximate bell shaped distribution, with
Mean() = 55
Variance (^2) = 36
Standard deviation() = 6
Solution(a)
We need to calculate percentage between 43 and 67 bottles to be sold in a day i.e. P(43<X<67) = ?
P(43<X<67) = P(X<67) - P(X<43)
Here we will use standard normal distribution, First we will calculate Z-score which can be calculated as
Z-score = (X-)/ = (43-55)/6 = -2
Z-score = (67-55)/6 = 2
From Z table we found p-value
P(43<X<67) = P(X<67) - P(X<43) = 0.9772 - 0.0227= 0.9545
So there is 95.45% between 43 and 67 bottles to be sold in a day.
Solution(b)
We need to calculate percentage of the time will the number of bottles sold more than 61 bottles
P(X>61) = 1 - P(X<=61)
Here we will use standard normal distribution, First we will calculate Z-score which can be calculated as
Z-score = (X-)/ = (61-55)/6 = 1
From Z table we found p-value
P(X>61) = 1 - P(X<=61) = 1 - 0.8413 = 0.1587
So there is 15.87% of the time will the number of bottles sold more than 61 bottles.
Solution(c)
If the supermarket begins each morning with a supply of 73 bottles how often will demand exceed the supply i.e. P(X>73) = 1 - P(X<=73)
Here we will use standard normal distribution, First we will calculate Z-score which can be calculated as
Z-score = (X-)/ = (73-55)/6 = 3
From Z table we found p-value
P(X>73) = 1 - P(X<=73) = 1 - 0.9987 = 0.0013
So there is 0.13% of the time will the number of bottles sold more than 73 bottles.