The state of virginia has implemented a standard of learning (sol) test that all public school students must pass before they can graduate from high school. a passing grade is 75. montgomery county high school administrators want to gauge how well their students might do on the sol test, but they don’t want to take the time to test the whole student population. instead, they selected 20 students at random and gave them the test. the results are as follows: 83 79 56 93 48 92 37 45 72 71 92 71 66 83 81 80 58 95 67 78 assume that sol test scores are normally distributed. a. compute the mean and standard deviation for these data. b. determine the probability that a student at the high school will pass the test. c. how many percent of students will receive a score between 75 and 95? d. what score will put a student in the bottom 15% in sol score among all students who take the test? e. what score will put a student in the top 2% in sol score among all students who take the test? g

a. Mean = 72.35

Standard deviation = 16.68

Mean =

and,

where, is mean of the distribution.

b. The passing grade is 75.

The students whose grade is less than 75 are fail

There are 10 students whose grade are more than 75.

Thus, Probability of pass students = 10 ÷ 20 = 0.5

c. There are 10 students that score between 75 and 95.

∴ Probability = 10 ÷ 20 = 0.5

d. The 15% of 20 = 3 students.

Thus the score will put a student in the bottom 15% = 48

e. Thus, the score will put a student in the top 2% = 95

range is infinite on both x and y axis positive and negative

step-by-step explanation:

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497