A certain company makes many many different puzzles all of t

A certain company makes many, many different puzzles, all of the same difficultly level. The times it takes Jack to assemble such puzzles are normally distributed with mean 500 minutes and standard deviation 30 minutes. Sam\'s times are normally distributed with mean 475 minutes and standard deviation 20 minutes. Find the following probabilities (be as accurate as our process allows).

1.Each are given a copy of the same puzzle. Find the probability that Jack will finish before Sam.

2.Find the probability that Jack\'s average time on 4 puzzles is less than Sam\'s average time on 4 puzzles.

Solution

Let J be the random variable denoting the time taken by Jack to assemble a puzzle.
Let S be the random variable denoting the time taken by Sam to assemble a puzzle.

As given, J follows N ( 500, 30^2 ) and S follows N ( 475 , 20^2 ) .

1. It\'s required to compute the probaility that J < S .

Pr ( J < S ) = Pr ( J - S < 0 )
Let T be a random variable denoted by J - S . Therefore, T follows Normal with mean 500 - 475 = 25.
Var ( T ) = Vat ( J - S ) = Var ( J ) + Var ( S ) . As, the random variables J and S are independent . The time taken by any one of them to assemble a puzzle is independent of the time taken by other to asseble.
Var ( T) = 30^2 + 20^2 = 1300
S.D ( T ) = 1300^(1/2)

Hence T follows Normal with mean 25 and variance 1300.
Pr ( T < 0 ) = Pr ( z < [- 25 / (1300^(1/2)) ] )
   = phi ( - .693375)
   = 1 - phi ( .693375)
= 1 - .755
   = .245

2. Let J_i be the random variable denoting the time taken by Jack on i^th puzzle.
  Let S_i be the random variable denoting the time taken by Sam on i^th puzzle. i = 1 to 4
   It\'s required to compute that Pr [ ( J₁ +.......... + J₄ / 4 ) < ( S₁ + ............ + S₄ / 4 ) ]
   i.e Pr ( T₁ + T₂ + .... + T₄ / 4 < 0 )
   The average of T_i^\'s where i = 1 to 4 follows Normal distribution with mean 25 and variance (1300 + 1300 + 1300 + 1300) / 16 = 1300 / 4

Therefore required probability is Pr [ z < ( - 50 / (1300^(1/2)) ]

Pr [ z < -1.386750491 ]

   phi ( - 1.386750491 )
   1 - phi ( 1.38675 )
   1 - .91774 = .08226