Sample Solution

Construct a table of the binomial probability distribution for this case:

Number of People Checking Weather | 0 | 1 | 2 | 3 | 4 | 5
Probability | 0.83| 0.17|0.028|0.0045|0.0006|0.00003

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Calculate the mean, variance, and standard deviation for this distribution:
The mean is 17%, or 0.17; the variance is 0.1509 and the standard deviation is 0.387567431 .

Calculate the probability that at least 3 people in the sample checked the weather that day:
The probability that at least 3 people in the sample checked the weather that day is 0.0045 (or 4.5%).

According to a recent infographic (NAMI: National Alliance on Mental Illness, n.d.), \”1 in 20 U.S. adults experience serious mental illness\” in the United States If a random sample of 20 students is selected from a large school district, construct a table of the binomial probability distribution for this case:
Number of Students Diagnosed with Serious Mental Illness | Probability
0 | (.95)^20 = .621341841                   1        20C1* (1-.95)^1*. 95^19= .235357068                     2                 20C2*(1-.95)^2*. 95^18= .096372793                                          3                        20C3*(1-.95) ^3 *. 95 ^17= .029078833                                        4                           20C4 * (1-.95) ^4 * 95 ^16= .007391177
5    20C5 * (1-.95 )5 * 95 ^15 = .001553469

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Calculate the mean, variance, and standard deviation for this distribution: The mean number of students out of twenty who will be diagnosed with seriousmental illness is 1;the varianceis0.3250andthestandarddeviationis057217967
If 3 ofthestudentswerediagnosedwithseriousmentailnesswouldthisbesignalabouttheschooldistricthavingproblemwithmentalhealthamongthestudentsbeing abovenationalscore? Yes,itwouldbeasignalthattheschooldistrictmayhaveaproblemwithhighratesofmentalillnesscomparedtonation

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