The average number of shoppers at a particular grocery store in one day is 506, and the standard deviation is 115. The number of shoppers is normally distributed. For a random day, what is the probability that there are less than 200 shoppers at the grocery store?

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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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

i think ur answer is   (3,-1)

first one is a: renaming a fraction. second one is c: de-c

step-by-step explanation:

my explanation is that these were the answers on edg : )