Problems 6 10 An investigator polls a representative sample

(Problems 6 - 10) An investigator polls a representative sample of common cold sufferers, asking them to estimate the number of hours of physical discomfort caused by their most recent cold. Their estimates approximate a normal curve with a mean of 83 hours and a standard deviation of 20 hour. 6. What percentage of sufferers estimate that their colds lasted for longer than forty - eight hours? 7. What percentage suffered for fewer than 61 hours? 8. What percentage suffered for between one and three days? 9. Inverse Normal Distribution: What is the estimated number of hours for the shortest - suffering 10 percent? 10. Inverse Normal Distribution: A medical researcher wishes to concentrate on the 20 percent who suffered the most. She will work only with those who estimate that they suffered for more than _____hours.

Solution

Normal Distribution
Mean ( u ) =83
Standard Deviation ( sd )=20
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
6.
P(X > 48) = (48-83)/20
= -35/20 = -1.75
= P ( Z >-1.75) From Standard Normal Table
= 0.9599 ~ 95.99% are longer than 48
7.
P(X < 61) = (61-83)/20
= -22/20= -1.1
= P ( Z <-1.1) From Standard Normal Table
= 0.1357 ~ 13.57% are lessar than 61                  
8.                  
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 24) = (24-83)/20
= -59/20 = -2.95
= P ( Z <-2.95) From Standard Normal Table
= 0.00159
P(X < 72) = (72-83)/20
= -11/20 = -0.55
= P ( Z <-0.55) From Standard Normal Table
= 0.29116
P(24 < X < 72) = 0.29116-0.00159 = 0.2896