Warm Up

• Section 4.21. 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.

x 3 6 8 11 12

P(x) .06 .19 .3 .21 .25

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.

x 3 6 8 11 12

P(x) .06 .19 .3 .21 .25

X – Profit $1 $2 -$2 Sum

P(X) 1/8 1/8 6/8 8/8 = 1

X*P(x) 1/8 2/8 -12/8 -9/8=-$1.13

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

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.

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