Information was collected that 10 percent of people admitted

Information was collected that 10 percent of people admitted to eating one or more pizza slices per day during football season. A random sample of 5 men are selected from this population. Let X= the number out of 5 who admitted eating pizza during football season.

1)Make a discrete probability distrubution for x.

2) P(X>2)

3) P (X > or equal to 2)

4)P (1<X<4)

Solution

1)

This is a binomial distribution with n = 5. Thus, the probability of x successes is

P(x) = nCx p^x (1-p)^(n-x)

P(x) = 5Cx (0.1)^x (1-0.1)^(5-x)

P(x) = 5Cx (0.1)^x (0.9)^(5-x) [ANSWER]

***************

2)

Note that P(more than x) = 1 - P(at most x).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    5      
p = the probability of a success =    0.1      
x = our critical value of successes =    2      
          
Then the cumulative probability of P(at most x) from a table/technology is          
          
P(at most   2   ) =    0.99144
          
Thus, the probability of at least   3   successes is  
          
P(more than   2   ) =    0.00856 [answer]

*******************

3)

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    5      
p = the probability of a success =    0.1      
x = our critical value of successes =    2      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   1   ) =    0.91854
          
Thus, the probability of at least   2   successes is  
          
P(at least   2   ) =    0.08146 [answer]

********************

4)

Here, 1<x<4, so 1 and 4 are not included, only 2 and 3.

Note that P(between x1 and x2) = P(at most x2) - P(at most x1 - 1)          
          
Here,          
          
x1 =    2      
x2 =    3      
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    5      
p = the probability of a success =    0.1      
          
Then          
          
P(at most    1   ) =    0.91854
P(at most    3   ) =    0.99954
          
Thus,          
          
P(between x1 and x2) =    0.081   [answer]