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Warm Up. Section 4.2 Explain why the following is not a valid probability distribution? Is the variable x discrete or continuous? Explain. 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. .
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Warm Up • Section 4.2 • Explain why the following is not a valid probability distribution? • Is the variable x discrete or continuous? Explain. • 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.
Warm Up • Section 4.2 • Explain why the following is not a valid probability distribution? Because sum of the probabilities is not 1. • Is the variable x discrete or continuous? Explain. I assume discrete because there are gaps. • 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.
Section 4.4 – Mean, Variance and Standard Deviation for the Binomial Distribution • Remember these values are what will probablyhappen, 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