Question

Owners of a minor league baseball team believe that a Normal model is useful in projecting the number of fans who will attend home games. They use a mean of 8500 fans and a standard deviation of 1500 fans. Using this model, answer these questions. a. Sketch and clearly label this Normal model, based on the 68-95-99.7 Rule. b. What proportion of the home games will have fewer than 6000 fans in attendance

Answer 1

Answer:

a) $\mu -\sigma = 8500 -1500 = 7000$

$\mu +\sigma = 8500 +1500 =10000$

$\mu -2*\sigma = 8500 -2*1500 = 5500$

$\mu +2*\sigma = 8500 +2*1500 =11500$

$\mu -3*\sigma = 8500 -3*1500 = 4000$

$\mu +3*\sigma = 8500 +3*1500 =13000$

On the figure attached we have the illustration for required showing the percentages and the limits for each case

b) $z= \frac{6000-8500}{1500}= -1.67$

And using the normal standard table or excel we got:

$P(Z<-1.67) = 0.047$

Step-by-step explanation:

For this case we have the following properties:

$\mu = 8500, \sigma = 1500$

Part a

And for this case from the empirical rule we know that within one deviation from the mean we have 68% of the values , within two deviation 95% and within 3 deviations 99.7%. So then we can find the limits like this:

$\mu -\sigma = 8500 -1500 = 7000$

$\mu +\sigma = 8500 +1500 =10000$

$\mu -2*\sigma = 8500 -2*1500 = 5500$

$\mu +2*\sigma = 8500 +2*1500 =11500$

$\mu -3*\sigma = 8500 -3*1500 = 4000$

$\mu +3*\sigma = 8500 +3*1500 =13000$

On the figure attached we have the illustration for required showing the percentages and the limits for each case

Part b

For this case we want to find this probability:

$P(X<6000)$

We can calculate the number of deviation below the mean with the z score formula given by:

$z = \frac{X -\mu}{\sigma}$

And replacing we got:

$z= \frac{6000-8500}{1500}= -1.67$

And using the normal standard table or excel we got:

$P(Z<-1.67) = 0.047$