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.

y=mx+b

step-by-step explanation:

m is slope, b is y-intercept

answer: d. $20

step-by-step explanation:

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