A population of values has a normal distribution with μ = 42.1 and σ = 84.5 .

Question

1. Find the probability that a single randomly selected value is between 30 and 58.4. Round your answer to four decimal places.   
2. Find the probability that a randomly selected sample of size  has a mean between 30 and 58.4. Round your answer to four decimal places.   

Answer 1

x is normal (42.1,8.5)

want to find P(30 < x < 58.4) transform inequality to range of standardized z score N(0,1):

P(30 < x < 58.4) = P((30-42)/8.5 < z < (58.4-42)/8.5) - simplify and find probability using table or technology

want to find P(30 < M < 58.4) transform inequality to range of standardized z score:

P(30 < M < 58.4) = P( (30-42)/(8.5/sqrt(195)) < z < ((58.4 - 42)/(8.5/sqrt(195)) - simplify and find probability using table or technology

Answer 2

if there's a misprint, and standard deviaton = 8.45, not 84.5

then 42.1-30 = 12.1, 12.1/8.45 = 1.4320 = z score

58.4-30 = 28.4, 28.4/8.45 = 3.3610

P(30<x<58.4) = .4239 + .5000 = .9239 = 92.39%

IF you really meant standard devation = 84.5

calculate z scores = (value-mean)/standard devation

use a z calculator or z tables

P(30<x<58.4) = .1316 + .0006 = .1322 = 13.22%