Question

(a) draw and label a sketch of the normal curve

(b) identify and shade the area of interest

(c) identify any formulas and values substituted

(d) identify the calculator command used and values entered into the calculator

(e) write your response as a decimal rounded to three places

The tread life of a particular brand of tire is a random variable best described by a normal distribution with a mean of 60,000 miles and a standard deviation of 2800 miles.

Use 100,000 or -100,000 as the upper or lower bound where necessary.

a.  What is the probability a particular tire of this brand will last longer than 57,200 miles?

b.  What is the probability a particular tire of this brand will last less than 58,000 miles?

c.  What is the probability a particular tire of this brand will last between 56,850 miles and 57,300 miles?

Answer 1

Define random variable X : tread life of a particular brand of tire

X is normally distributed with mean = = 60000 and standard deviation = = 2800

a)

Here we have to find P(X > 57200)

   where z is standard normal variable.

  

= P(z > -1)

= 1 - P(z < -1)

= 1 - 0.1587 (From statistical table of negative z values)

= 0.8413

The probability a particular tire of this brand will last longer than 57,200 miles is 0.8413

b)

Here we have to find P(X < 58000)

   where z is standard normal variable.

  

= P(z < -0.71)    (Round to 2 decimal)

= 0.2389 (From statistical table of negative z values)

The probability a particular tire of this brand will last less than 58,000 miles is 0.2389

c)

Here we have to find P(56850 < X < 57300)

     

= P(-1.13 < z < -0.96) (Round to 2 decimal)

= P(z < -0.96) - P(z < -1.13)

= 0.1675 - 0.1303 (From statistical table of negative z values)

= 0.0372

The probability a particular tire of this brand will last between 56,850 miles and 57,300 miles is 0.0372