A normal distribution has a mean of 10 and a standard deviation of 1. What is the probability of selecting a number between 8 and 12? Leave the answer in decimal form to 3 decimal places

95.44%

Step-by-step explanation:

What we must do is calculate the z value for each value and thus find what percentage each represents and the subtraction would be the percentage between those two values.

We have that z is equal to:

z = (x - m) / (sd)

x is the value to evaluate, m the mean, sd the standard deviation

So for 8 we have:

z = (8 - 10) / (1)

z = -2

and this value represents 0.0228

for 12 we have:

z = (12 - 10) / (1)

z = 2

and this value represents 0.9772

we subtract:

0.9772 - 0.0228 = 0.9544

Which means that it represents 95.44%

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

p

step-by-step explanation: