The triglyceride levels of residents of an assisted living facility are recordered the levels are normally distributed with a mean of 200 and a standard deviation of 50. if a resident is randomly selected, find the probability of a level that is between 200 and 275

Step-by-step explanation:

P(200<X<275)

P(200-200/50 <X< 275-200/50)

P(0/50 <X< 75/50)

P(0<X<1.5)

P(X<1.5) -P(X>0)

0.9332-0.5 = 0.4332

The probability is 0.4332.

Explanation:

Mean = μ = 200

Standard deviation = σ = 50

We are required to find the probability that a randomly selected resident has a triglyceride level between 200 and 275. That is,

P (200 < X < 275) = ?

We are further informed that the triglyceride levels are normally distributed. We, therefore, convert it into standard normal distribution (Z-distribution), and compute it as follows.

P (200 < X < 275)

= P [(200 - 200)/50 < (X - μ)/σ < (275 - 200)/50]

= P (0 < Z < 1.5)

= P(Z < 1.5) - P(Z < 0)

= 0.9332 - 0.5

= 0.4332

Therefore, the probability that a randomly selected resident has a triglyceride level between 200 and 275 is 0.4332.

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1