Q. Of a large group of men, 5% are under 58 inches and 40% are between 58 and 65 inches. Assuming a normal distribution find the mean height and standard deviation.
please answer with solutions
Answers
Given : Of a large group of men,
5% are under 58 inches and
40% are between 58 and 65 inches.
A normal distribution
To find : the mean height and standard deviation.
Solution:
5% are under 58 inches
40% are between 58 and 65 inches.
=> 5 + 40 = 45 % under 65 inches.
Mean = M inches
SD = S inches ( SD = Standard Deviation )
Z score = ( Value - Mean)/SD
5% are under 58 inches
Z score for 5 % = -2.575
-2.575 = ( 58 - M)/SD
45 % under 65 inches.
Z score for 45 % = -0.125
-0.125 = ( 65 - M)/SD
20.6 = (58 - M)/(65 - M)
=> 1339 - 20.6M = 58 - M
=> 19.6M = 1281
=> M = 65.357
Substituting in any one
S = 2.857
Mean height = 65.357 inches
SD = 2.857 inches
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Answer:
Mean Height:- 65.357 inches
Standard Deviation:- 2.857 inches
Step-by-step explanation:
Given: 5% are under height 58 inches and 40%are in between 58 and 65 inches.
therefore,
total % of men who are under 65 inches are
= (5+40)%
=45%
Now,
Zscore = (value-mean)/SD [SD = standard deviation]
Zscore for 5% under 58 inches
=-2.575
therefore,
-2.575=(58-M)/SD
Zscore for 45% under 65 inches
=-0.125
therefore,
-0.125=(65-M)/SD
20.6=(58-M)/(65-M)
=>1339-20.6M = 58-M
=> 19.6M = 1281
=>M = 65.357 inches
Substituting in any one
we get ,
SD =2.857 inches.