Every day, 17% of the U.S. population with Internet access checks the weather in that way. For a sample of 5 people taken on any given day:
Construct a table of the binomial probability distribution for this case.
View this information using a histogram.
Calculate the mean, variance, and standard deviation for this distribution.
Calculate the probability that at least 3 people in the sample checked the weather that day.
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.
View this information using a histogram.
Calculate the mean, variance, and standard deviation for this distribution.
If 3 of the students were diagnosed with serious mental illness, would this be a signal about the school district having a problem with mental health among the students being above the national score?+1200 words
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
View this information using a histogram:
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
View this information using a histogram:
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