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1.A normal distribution has a mean of 102 and a standard deviation of 2.9 

what is the Z-score of 105? 

Round to the nearest hundredth in the box 

2. The age of members of an online gaming group are normally distributed with a mean of 23 and a standard deviation of 4 years. 

There are 300 members in the group about how many members are expected to be between 15 years old and 31 years old? 

A. 204

B.264

C.270

D.285

 Apr 12, 2019
 #1
avatar+18293 
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(105-102)/2.9 = + 1.03 S.D. above the mean

 

 

 

31  is 2 standard deviations ABOVE    z score  corresponds to  fraction .9772

15  2 standard deviations BELOW      z score corresponds to fraction .0228

 

.9772 - .0228 = Fraction of total between 15 and 31 = .9544

.9544 (300) = ~ 285 members

 Apr 12, 2019

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