Answer to Question #275882 in Statistics and Probability for Shen

Question

Noel, a 35 year old office worker, has been struggling with his weight for a long time. He is looking to start dating again, so earlier this year he decided to join a gym. A few days ago he noted that his average weight for the past 6 months is 220.6lbs and the standard deviation is 96. He is hoping that his weight will be between 160lbs and 180lbs by the end of January.

What is the probability that he will achieve the weight he desires? Interpret your result 

Answer 1

P(160<x<180)

$Z=\frac{(x-\mu)}{\sigma}$

P(x<180)-P(x<160)

$=\phi(z2)-\phi(z1)$

$$=\phi(\frac{180-220.6}{96})-\phi(\frac{160-220.6}{96})$$

=0.33618-0.26394

=0.07224