Math Home l My Classes Assessment Due in 3 hours, 4 minutes. Due Sun 02/25/2018
ID: 3050142 • Letter: M
Question
Math Home l My Classes Assessment Due in 3 hours, 4 minutes. Due Sun 02/25/2018 11:59 p Show Intro/Instructios A population of values has a normal distribution with size n = 17 185.3 and 9.8. You intend to draw a random sample or Find the probability that a single randomly selected value is greater than 178.2 P(X 178.2)- Find the probability that a sample of size n P(M> 178.2) 17 s randomly selected with a mean greater than i 782. Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact -scores or z-scores rounded to 3 decimal places are accepted Get help: Video Points possible: 2 This is attempt 1 of 3 Post this question to forum Submit 8 0.Explanation / Answer
Solution:
Given information = 185.3 and = 9.8 we have:
P(X > 178.2 ) = P( X/ > 178.2185.3/9.8)
= P(Z > 0.72 )
= 0.7642
The probability that a sample of size n = 17 is randomly selected with a mean greater than 178.2.
P(M > 178.2 ) = P(X/(/n) > 178.2185.3/(9.8/17)
= P(Z > -2.9871 )
= P ( Z<2.9871 )
= 0.9986