Question

The total duration of baseball games in the major league in the 2011 season is uniformly distributed between 447 hours and 521 hours inclusive.

1. Find $a$and $b$and describe what they represent.
2. Write the distribution.
3. Find the mean and the standard deviation.
4. What is the probability that the duration of games for a team for the 2011 season is between 480 and 500 hours?
5. What is the 65th percentile for the duration of games for a team for the 2011 season?

Short answer

1. $a=447$and $b=521$and it indicates the minimum and maximum duration of the baseball games in the major league in the 2011 season.
2. The distribution is uniform distribution.
3. The value of mean is $484$and the standard deviation is $21.36$
   
4. The probability that the duration of games for a team for the 2011 season is between 480 and 500 hours is$0.3456$.
5. The value of the ${65}^{th}$percentile is $492.43$.

Step-by-step solution

Part (a) Step 1: Given Information 

Given in the question that, the total duration of baseball games in the major league in the 2011 season is uniformly distributed between 447 hours and 521 hours inclusive.

We have to find$a$and$b$and describe what they represent.

Part (a) Step 2: Solution 

The total duration of the baseball games is uniformly distributed.

Therefore, $X~U(447,521)$

As per the given information the value of $a$and $b$is $447$and $521$.

The value of $a$indicates the minimum total duration of the baseball is $447$hours and $b$indicates the maximum duration of the baseball game is $521$hours in the major league in the 2011 season.$447$

Part (b) Step 1: Given Information

Given in the question that, the total duration of baseball games in the major league in the 2011 season is uniformly distributed between 447 hours and 521 hours inclusive.

We need to write the distribution

Part (b) Step 2: Solution 

The overall duration of baseball games is uniformly distributed, according to the statistics presented. As a result, the distribution is uniform distribution

Part (c) Step 1: Given Information 

Given in the question that, the total duration of baseball games in the major league in the 2011 season is uniformly distributed between 447 hours and 521 hours inclusive.

We have to find the mean and the standard deviation.

Part (c) Step 2: Calculate the value of mean 

The mean of uniform distribution can be calculated as,

$\mu =\frac{a+b}{2}$

Here,

$a=447$

$b=521$

So,

$\mu =\frac{447+521}{2}$

$=\frac{968}{2}$

$=484$

Part (c) Step 3: Calculate the standard deviation 

The formula which is used to find the standard deviation is:

$\sigma =\sqrt{\frac{(b-a{)}^2}{12}}$

Place the provided values in the equation.

Therefore,

$\sigma =\sqrt{\frac{(521-447{)}^2}{12}}$

$=\sqrt{\frac{5476}{12}}$

$=\sqrt{456.33}$

$=21.36$

Part (d) Step 1: Given Information 

Given in the question that, the total duration of baseball games in the major league in the 2011 season is uniformly distributed between 447 hours and 521 hours inclusive.

We need to find what is the probability that the duration of games for a team for the 2011 season is between 480 and 500 hours?

Part (d) Step 2: Solution 

The required probability can calculated as

$P(480<x<500)=P(x<500)-P(x<480)$

$=P\left(\frac{x-\mu }{\sigma }<\frac{500-\mu }{\sigma }\right)-P\left(\frac{x-\mu }{\sigma }<\frac{480-\mu }{\sigma }\right)$

$=P\left(Z<\frac{500-484}{21.36}\right)-P\left(Z<\frac{480-484}{21.36}\right)$

$=P(Z<0.7491)-P(Z<-0.1873)$

$=0.7731-0.4275\left\{\begin{array}{l}\text{From standard}\\ \text{normal deviation}\end{array}\right\}$

$=0.3456$

Part (e) Step 1: Given Information 

Given in the question that, the total duration of baseball games in the major league in the 2011 season is uniformly distributed between 447 hours and 521 hours inclusive.

We have to what is the 65th percentile for the duration of games for a team for the 2011 season.

Part (e) Step 2: Solution 

Let's calculate the ${65}^{th}$percentile as below

$P(x<k)=0.65$

$P\left(\frac{x-\mu }{\sigma }<\frac{k-\mu }{\sigma }\right)=0.65$

$P\left(Z<\frac{k-484}{21.36}\right)=0.65$

$\frac{k-484}{21.36}=0.3853\left\{\begin{array}{l}\text{From standard}\\ \text{normal table}\end{array}\right\}$

$k-484=0.3853\times 21.36$

$k=8.23+484$

$k=492.23$