Question

A. A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 162.4-cm and a standard deviation of 0.6-cm. For shipment, 16 steel rods are bundled together.
Find the probability that the average length of a randomly selected bundle of steel rods is less than 162.6-cm.
P(¯xx¯ < 162.6-cm) =

B. A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 246.8-cm and a standard deviation of 1.1-cm. For shipment, 24 steel rods are bundled together.
Find the probability that the average length of a randomly selected bundle of steel rods is greater than 247.3-cm.
P(¯xx¯ > 247.3-cm) =

C. Business Weekly conducted a survey of graduates from 30 top MBA programs. On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is 149000 dollars. Assume the standard deviation is 39000 dollars. Suppose you take a simple random sample of 77 graduates.
Do not use probability tables to find the probabilities below as they may not be accurate enough.
Find the probability that a single randomly selected salary is at least 144000 dollars.
P(X > 144000) =
Find the probability that a sample of size n=77n=77 is randomly selected with a mean that is at least 144000 dollars.
P(¯xx¯ > 144000) =

Answer 1

Solution :

a ) Given that,

mean = = 162.4

standard deviation = =0.6

n = 16

= 162.4

  =  ( /n) = (0.6 / 16 ) =0.15

P (   < 162.6 )

P ( - /) < (162.6 - 162.4 /0.15)

P ( z < 0.2 /0.15 )

P ( z < 1.33 )

Using z table

=0.9082

Probability = 0.9082

b ) Given that,

mean = = 246.8

standard deviation = = 1.1

n = 24

= 246.8

  =  ( /n) = ( 1.1 / 24 ) = 0.2245

P (   >   247.3 )

= 1 - P (   < 247.3 )

= 1 - P ( - /) < ( 247.3 - 246.8 /0.2245)

= 1 - P ( z < 0.5 /0.2245 )

= 1 - P ( z < 2.23 )

Using z table

= 1 - 0.9871

= 0.0129

Probability = 0.0129

C ) Given that,

mean = = 149000

standard deviation = = 39000

P ( X >    144000 )

= 1 - P ( X <  144000 )

= 1 - P ( X - /) < ( 144000 -149000 /39000 )

= 1 - P ( z < - 5000 / 39000 )

= 1 - P ( z < 0.13 )

Using z table

= 1 - 0.4483

= 0.5517

Probability = 0.5517

n = 77

= 149000

  =  ( /n) = ( 39000 / 77 ) = 4444.62

P ( >  144000 )

= 1 - P ( <  144000 )

= 1 - P ( - /) < ( 144000 -149000  / 4444.62 )

= 1 - P ( z < - 5000 /4444.62 )

= 1 - P ( z < - 1.12 )

Using z table

= 1 - 0.1314

= 0.8686

Probability = 0.8686