M/S Talwarkar Fitness Center as a part of their physical fitness programme, want their 500 clients to lose their excess weight, through a programme developed by Dr. Sharma. For that programme they claim that the average weight lost by their clients is 10 kg with 2 kg as standard deviation. If weight lost are normally distributed, find
I) The number of clients losing more than 13.5 kg.
Ii) The minimum weight lost by 90% clients.
Iii) Find the number of clients losing weight between 5 kg and 11 kg.
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Answers
Given : 500 clients
average weight lost by their clients is 10 kg
standard deviation 2 kg
To Find : I) The number of clients losing more than 13.5 kg.
Ii) The minimum weight lost by 90% clients.
Iii) Find the number of clients losing weight between 5 kg and 11 kg.
Solution:
Z score = ( Value - Mean) / SD
The number of clients losing more than 13.5 kg.
=> Z score = ( 13.5 - 10)/2 = 1.75
Z score 1.75 means 0.9599
Hence losing more than 13.5 kg. = 1 - 0.9599 = 0.0401
losing more than 13.5 kg. = 500 x 0.0401 = 20
20 clients losing more than 13.5 kg.
The minimum weight lost by 90% clients.
Hence 10 % lost below that = 0.1
z score = 1.282
-1.282 = ( Value - 10) /2
=> -2.564 = Value - 10
=> Value = 7.436 kg
The minimum weight lost by 90% clients is 7.436 kg
number of clients losing weight between 5 kg and 11 kg.
Z score for 5 = (5 - 10)/2 = - 2.5 => 0.62 %
Z score for 11 = (11 - 10)/2 = 0.5 => 69.15 %
Between = 69.15 - 0.62 = 68.53 %
Number of clients = ( 68.53 /100) x 500 = 342.65 = 343
343 clients losing weight between 5 kg and 11 kg.
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