For a random day with the standard deviation, the probability that there are less than 200 shoppers on a random day, is 0.0039.
The average total number of shoppers on a grocery store in 1 day is 506; the standard deviation is 115.
The number of shoppers is normally distributed.
Let, X be the random variable denoting the number of shoppers on a random day.
Then, X follows normal with mean 506 and standard deviation of 115.
Then, we can say that,
Z=(X-506)/115 follows standard normal with mean 0 and standard deviation of 1.
We have to find
P(X<200)
[tex]=P(Z < \frac{200-506}{115})[/tex]
=P(Z<-2.66)
Where, Z is the standard normal variate.
ρ = -0.266
Where, ρ is the distribution function of the standard normal variate.
From the standard normal table, this becomes
=0.0039
For a random day, the probability that there are less than 200 shoppers on a random day, is 0.0039.
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