Question

A population of values has a normal distribution with μ=115.6 and σ=46.5. You intend to draw a random sample of size n=183.

a. Find the probability that a single randomly selected value is between 118.7 and 126.6.
P(118.7 < X < 126.6) =

b. Find the probability that a sample of size n=183 is randomly selected with a mean between 118.7 and 126.6.
P(118.7 < ¯xx¯ < 126.6) =

Enter your answers as numbers accurate to 4 decimal places.

Answer 1

Formulas: 1) Ifu, o are the mean and standard deviation of the population. If we take n samples out of the population then sample mean is ug = u and sample standard deviation is og u to = Facts : 1) If X and Z are the normal and standard normal distibuted random variables respectively. X-μ X-interval can be converted to Z-interval by using Z o Where u = mean(X) o = standard deviation(X) Let X = value of a population. Given that X follows normal distribution with mean(u) = 115.6 and standard deviation(o) = 46.5

a) Now we have to find P(118.7 < X < 126.6). P(118.7 < X < 126.6) = P(118.7 – 115.6 <X – u < 126.6 – 115.6) X- Р 118.7-115.6 46.5 126.6-115.6 46.5 = P(0.07 < Z <0.24) = P(Z < 0.24) - P(Z < 0.07) = 0.5948 – 0.5279 = 0.0669 .: P(118.7 < X < 126.6) = 0.0669

b) X is sample mean random variable of 183 samples. Sample size = n = 183 Mg = u = 115.6 46.5 ох 3.4374 Vn 183 Now we have to find P(118.7 < X < 126.6). P(118.7 < X < 126.6) = P(118.7 – 115.6 < X - už <126.6 – 115.6) 118.7-115.6 PE x-ux 126.6-115.6 5.6-115.5 3.4374 ах 3.4374 - P(0.90 < 3 < 3.20) P(Z < 3.20) - P(Z < 0.90) = 0.9993 – 0.8159 = 0.1834 :: P(118.7 < X < 126.6) = 0.1834

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