Question

1. A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 243.6-cm and a standard deviation of 2.2-cm. For shipment, 22 steel rods are bundled together.

Find the probability that the average length of a randomly selected bundle of steel rods is between 242.8-cm and 244.3-cm.
P(242.8-cm < ¯x¯ < 244.3-cm) =

2. CNNBC recently reported that the mean annual cost of auto insurance is 1031 dollars. Assume the standard deviation is 134 dollars. You will use a simple random sample of 107 auto insurance policies.

Find the probability that a single randomly selected policy has a mean value between 1011.6 and 1021.9 dollars.
P(1011.6 < X < 1021.9) =

Find the probability that a random sample of size n=107n=107 has a mean value between 1011.6 and 1021.9 dollars.
P(1011.6 < x < 1021.9) =

Answer 1

Solution :

Given that,

mean = = 243.6

standard deviation = = 2.2

n = 22

= 243.6

= / n = 2.2 / 22 = 0.4690

P( 242.8 < < 244.3 )  

= P[( 242.8 - 243.6 ) / 0.4690 < ( - ) / < ( 244.3 - 243.6 ) / 0.4690 )]

= P( -1.71 < Z < 1.49)

= P(Z < 1.49 ) - P(Z < -1.71 )

Using z table,  

= 0.9319 - 0.0436

= 0.8883

Probability = 0.8883

2.

Solution :

Given that ,

mean = = 1031

standard deviation = = 134

P( 1011.6 < x < 1021.9 ) = P[( 1011.6 - 1031 ) /134 ) < (x - ) /  < ( 1021.9 - 1031 ) / 134 ) ]

= P( -0.14 < z < -0.07 )

= P(z < -0.07 ) - P(z < -0.14 )

Using z table,

= 0.4721 - 0.4443

= 0.0278

Probability = 0.0278

n = 107

= 1031

= / n = 134 / 107 = 12.9542

P(1011.6 < < 1021.9 )  

= P[( 1011.6 - 1031 ) / 12.9542 < ( - ) / < ( 1021.9 - 1031 ) / 12.9542 )]

= P( -1.50 < Z < -0.70 )

= P(Z < -0.70 ) - P(Z < -1.50 )

Using z table,  

= 0.2420 - 0.0668  

= 0.1752

Probability = 0.1752