Areas under The normal Curve

Question

The mean of 200 numbers is 52 with a standard deviation of 7, what is the probability of getting a number between 50and 72?

Write your answer in four decimal places.

Answer 1

Using on line assets instead of tables in books, or calculator functions [ex. TI-83. nmcdf(lb,ub,mean,sd) ]

https://onlinestatbook.com/2/calculators/normal_dist.html

Specify Parameters:

Mean  52

  SD  7

 Set limits

Between  50 and 72

 

Results: Area (probability) = .6103 or using TI-83 nmcdf(50,72,52,7) = 0.610314025076. (.6103)

NOTE: don't let extra information fool you. The size of the sample is totally immaterial as we are NOT testing means but distribution. SD is used for the calculations, not SD of mean (otherwise known as Standard Error)

Alternate calculation.

Find Z test for lower and upper bounds, then subtract left sided p value for lower Z value from upper Z value

Zlower = (50-52)/7 or -.28571428 P(<Zlower) = .3875486

Zupper = (72-52)/7 or 2.587143. P(<Zupper) = .9978626

P(<Zupper) - P(<Zlower) =. .9978626 -.3875486 = .610314. (rounded to 0.6103)

note: always do calculations at least 2 decimal places more than required answer accuracy to avoid rounding errors.

Answer 2

use an online z score calculator. google it and several come up

(72-52)/7 = 30/7 = 2.85714 = z score

P(52<x<72) = .49786

(52-50)/7 = 2/7 = 0.28571

Pr(50<x<52) =0.11245

Pr(50<x <72) = Pr(50<x<52) + Pr(52<x<72) = .11245 + .49786 = 0.61031 =0.6103 rounded off