Respuesta :
Answer:
The probability is 0.003907
Step-by-step explanation:
We start by calculating the z-score
Mathematically;
z-score = (x-mean)/SD
mean = 506, SD = 115 , x = 200
z-score = (200-506)/115 = -2.66
The probability we want to calculate is;
P( x < -2.66)
we can use the standard normal distribution for this;
P(x < -2.66) is ) = 0.003907
The probability that there are less than 200 shoppers at the grocery store is; 0.003907
We are given;
Population mean; μ = 506
Standard deviation; σ = 115
Sample mean; x' = 200
Since the number of shoppers is normally distributed, let us find the z-score of this distribution from the formula;
z = (x' - μ)/σ
Plugging in the relevant values gives;
z = (200 - 506)/115
z = -2.66
Now, we want to find the probability that there are less than 200 shoppers at the grocery store. This is; P(x < z)
From the z-score distribution table, we have;
P(x < -2.66) = 0.003907
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