Question

1. A machine precision-cuts tubing such that μ = 38.00 in. and σ = 0.03 in. If the lengths are normally distributed, find the probability that a piece of tubing selected at random measures as follows. (Round your answers to four decimal places.)

   (a) greater than 38.05 in.
   
   
   (b) less than 38.045 in.
   
   
   (c) between 37.95 and 38.05 in.

2. A certain model of automobile has its gas mileage (in miles per gallon, or mpg) normally distributed, with a mean of 32 mpg and a standard deviation of 4 mpg. Find the probability that a car selected at random has the following gas mileages. (Round your answers to four decimal places.)

   (a) less than 30 mpg
   
   
   (b) greater than 38 mpg
   
   
   (c) between 26 and 38 mpg

Answer 1

Solution :

Given that ,

1)

a)

P(x > 38.05) = 1 - P(x < 38.05)

= 1 - P((x - ) / < (38.05 - 38) / 0.03)

= 1 - P(z < 1.67)

= 1 - 0.9525 Using standard normal table.

= 0.0475

Probability = 0.0475

b)

P(x < 38.045) = P((x - ) / < (38.045 - 38) / 0.03)

= P(z < 1.5)

= 0.9332 Using standard normal table,

Probability = 0.9332

c)

P( 37.95 < x < 38.05) = P((37.95 - 38)/ 0.05) < (x - ) /  < (38.05 - 38) / 0.05) )

= P(-1.67 < z < 1.67)

= P(z < 1.67) - P(z < -1.67)

= 0.9525 - 0.0475 Using standard normal table,  

Probability = 0.9050

2)

a)

P(x < 30) = P((x - ) / < (30 - 32) / 4)

= P(z < -0.5)

= 0.3085 Using standard normal table,

Probability = 0.3085

b)

P(x > 38) = 1 - P(x < 38)

= 1 - P((x - ) / < (38 - 32) / 4)

= 1 - P(z < 1.5)

= 1 - 0.9332 Using standard normal table.

= 0.0668

Probability = 0.0668

c)

P(26 < x < 38) = P((26 - 32)/ 4) < (x - ) /  < (38 - 32) / 4) )

= P(-1.5 < z < 1.5)

= P(z <1.5 ) - P(z < -1.5)

= 0.9332 - 0.0668 Using standard normal table,  

Probability = 0.8664