Answer:
a) 0.5768 = 57.68% probability that the shop sells at least 3 in a week.
b) 0.988 = 98.8% probability that the shop sells at most 7 in a week.
c) 0.0104 = 1.04% probability that the shop sells more than 20 in a month.
Step-by-step explanation:
For questions a and b, the Poisson distribution is used, while for question c, the normal approximation is used.
Poisson distribution:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
In which
x is the number of successes
e = 2.71828 is the Euler number
is the mean in the given interval.
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
The Poisson distribution can be approximated to the normal with , if
.
Poisson variable with the mean 3
This means that .
(a) At least 3 in a week.
This is . So
In which:
Then
So
0.5768 = 57.68% probability that the shop sells at least 3 in a week.
(b) At most 7 in a week.
This is:
In which
Then
0.988 = 98.8% probability that the shop sells at most 7 in a week.
(c) More than 20 in a month (4 weeks).
4 weeks, so:
The probability, using continuity correction, is P(X > 20 + 0.5) = P(X > 20.5), which is 1 subtracted by the p-value of Z when X = 20.5.
has a p-value of 0.9896.
1 - 0.9896 = 0.0104
0.0104 = 1.04% probability that the shop sells more than 20 in a month.