Question

1. The weights (in pounds) of adult (18 years or older) males in the US are approximately normally

   distributed with mean μ = 187 and standard deviation σ = 21.2.

   1. What is the probability that a randomly chosen adult male in the US will weigh at most 212?

   2. What is the probability that the mean weight of a random sample of 23 adult males in the US will be at least 177 pounds?

Answer 1

Solution :

Given that,

mean = = 187

standard deviation = =21.2   

P(X<212 ) = P[(X- ) / < (212-187) /21.2 ]

= P(z <1.18 )

Using z table

= 0.8810

probability=0.8810

(B) n=23

= =187

= / n = 21.2 / 23 = 4.4205

P( >177 ) = 1 - P( <177 )

= 1 - P[( - ) / < (177-187) / 4.4205]

= 1 - P(z < -2.26)

Using z table

= 1 - 0.0119

= 0.9881

probability= 0.9881