Question

A Starbucks coffee shop serves an average of 3500 customers per day, with a standard deviation of 250. What is the probability they will serve between 3300 and 4000 customers on any given day

Answer 1

Answer:

$P(3300<X<4000)=P(\frac{3300-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{4000-\mu}{\sigma})=P(\frac{3300-3500}{250}<Z<\frac{4000-3500}{250})=P(-0.8<Z<2)$  

And we can find this probability with this difference:  

$P(-0.8<Z<2)=P(Z<2)-P(Z<-0.8)=0.97725-0.2119=0.765$  

Step-by-step explanation:

Previous concepts  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem  

Let X the random variable that represent the amount of cofee shops of a population, and for this case we know the distribution for X is given by:  

$X \sim N(3500,250)$  

Where $\mu=3500$ and $\sigma=250$  

We are interested on this probability  

$P(3300<X<4000)$  

And the best way to solve this problem is using the normal standard distribution and the z score given by:  

$z=\frac{x-\mu}{\sigma}$  

If we apply this formula to our probability we got this:  

$P(3300<X<4000)=P(\frac{3300-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{4000-\mu}{\sigma})=P(\frac{3300-3500}{250}<Z<\frac{4000-3500}{250})=P(-0.8<Z<2)$  

And we can find this probability with this difference:  

$P(-0.8<Z<2)=P(Z<2)-P(Z<-0.8)=0.97725-0.2119=0.765$  

Answer 2

Answer:

Probability they will serve between 3300 and 4000 customers on any given day is 0.7654.

Step-by-step explanation:

We are given that a Starbucks coffee shop serves an average of 3500 customers per day, with a standard deviation of 250.

Assuming the data follows normal distribution.

Firstly, Let X = No. of customers served by Starbucks coffee shop

The z score probability distribution for is given by;

         Z = $\frac{ X - \mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 3500

            $\sigma$ = standard deviation = 250

Probability that they will serve between 3300 and 4000 customers on any given day is given by = P(3300 < X < 4000) = P(X < 4000) - P(X $\leq$ 3300)

          P(X < 4000) = P( $\frac{ X - \mu}{\sigma}$ < $\frac{4000-3500}{250}$ ) = P(Z < 2) = 0.97725

          P(X $\leq$ 3300) = P( $\frac{ X - \mu}{\sigma}$ $\leq$ $\frac{3300-3500}{250}$) = P(Z $\leq$ -0.8) = 1 - P(Z < 0.8)

                                                                 = 1 - 0.78814 = 0.21186

Therefore, P(3300 < X < 4000) = 0.97725 - 0.21186 = 0.7654

Hence, probability that they will serve between 3300 and 4000 customers on any given day is 0.7654.