P 1452 1452 7 Find the probability that c Six MMs are brown 52P6086 460146

# P 1452 1452 7 find the probability that c six mms are

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P = 14/52 14/52 = 7 Find the probability that c.) Six M&M’s are brown. 52P6(0.86)^ 46(0.14)^6 = 20358520*0.0010*0.0000 = 0.1487 d.) Twenty-five M&M’s are brown. 52P25(0.86)^27(0.14)^25 = 0.0000 e.) All of the M&M’s are brown. 52P52(0.86)^0(0.14)^52 = 1*0.0170*1 = 0.017 f.) Would it be unusual for a package to have only brown M&M’s? If this were to happen, what would you think is the reason? Yes because it would be very unusual to only have brown M&Ms in the package. The reasoning this might happen is because a package of milk chocolate M&Ms contains 14% or 7 M&Ms that are brown total. This means that all of the other five colors; yellow, red, orange, green, and blue, are not in the packet. 5.3.4 Approximately 10% of all people are left-handed. Consider a grouping of fifteen people. a.) State the random variable. X = number of people left handed out of 15 people b.) Write the probability distribution. 15Px(0.10)^x(0.90)^15-x
c.) Draw a histogram. d.) Describe the shape of the histogram. e.) Find the mean. Mean = 1.5 μ = np μ = 15×0.10 = 1.5 f.) Find the variance. Variance = 1.35 0^2=npq Q=1-p Q = 0.90 =15*0.10*0.90 =1.35 g.) Find the standard deviation. Standard deviation = 1.162

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