1. Warm Up Section 4.2 1.Explain why the following is not a valid probability distribution? 2.Is the variable x discrete or continuous? Explain. 3.Find the expected profit of a game that costs $2 to play. You flip a coin three times, 3 heads wins $3, 3 tails wins $4, every thing else wins nothing. x3681112 P(x).06.19.3.21.25

2. Warm Up Section 4.2 1.Explain why the following is not a valid probability distribution? Because sum of the probabilities is not 1. 2.Is the variable x discrete or continuous? Explain. I assume discrete because there are gaps. 3.Find the expected profit of a game that costs $2 to play. You flip a coin three times, 3 heads wins $3, 3 tails wins $4, every thing else wins nothing. x3681112 P(x).06.19.3.21.25 X – Profit$1$2-$2Sum P(X)1/8 6/88/8 = 1 X*P(x)1/82/8-12/8-9/8=-$1.13

3. Section 4.4 – Mean, Variance and Standard Deviation for the Binomial Distribution Remember these values are what will probably happen, not what has happened. – Mean:  = n·p – Variance:  2 = n·p·q – Standard Deviation:  =  2 = n·p·q These values can be used to determine usual values. – Minimum usual value =  - 2  – Maximum usual value =  + 2 

4. Example A report states 40% of cell phone users have Verizon as their carrier. If 50 cell phone users are selected, find the mean, variance and standard deviation, assuming it’s binomial. Would it unusual to find that 28 of the 50 people use Verizon? Show work.

5. Example A report states 40% of cell phone users have Verizon as their carrier. If 50 cell phone users are selected, find the mean, variance and standard deviation. –  =.4*50 = 20  2 =.4*50*.6 = 12 –  = √12 = 3.464  3.4 Would it unusual to find that 28 of the 50 people use Verizon? Show work. – Max usual value = 20 + 2(3.464) = 26.928 – Yes it is unusual since it falls above the maximum unusual value of 26.928