2929Suppose a random number generator is used to generate tw

29.29.Suppose a random number generator is used to generate two numbers between 0 and 1. When these numbers are added together, their sum is Y. Note Y can take any value between 0 and 2. The density curve of Y is the triangle shown in the following figure:

30.A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 180.9-cm and a standard deviation of 2.2-cm.

A steel rod is chosen at random from all those produced by the company. What is the probability that the length of this rod is greater than 185.7 cm.


Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

31.The Normal distribution with mean ? = 6.8 and standard deviation ? = 1.6 is a good description of the Iowa Test vocabulary scores of seventh-grade students in Gary, Indiana. This is a continuous probability model for the score of a randomly chosen student. The density curve is shown below. Call the score of a randomly chosen student X for short.

What is the probability of this a student chosen at random has a score of 7.1 or higher?
% Round to the nearest 0.01%

33.A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 249.8-cm and a standard deviation of 2.3-cm. Suppose a rod is chosen at random from all the rods produced by the company. There is a 22% probability that the rod is longer than:

Enter your answer as a number accurate to 1 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Box 1: Enter your answer as a whole or decimal number. Examples: 3, -4, 5.5
Enter DNE for Does Not Exist, oo for Infinity

Solution

29.
probability that Y is less than 0.4 is
area under the curve between x = 0 and x = 0.4

at x=0, y=0
at x=0.4, y=0.4

required probability = (1/2)*0.4*0.4
= 0.08
= 8%

30.
population mean,u = 180.9
standard deviation,sigma = 2.2

P(X> 185.7 ) = P(Z> ((185.7-180.9) / 2.2)
= P(Z> 2.182 )
= 1 - P(Z<2.182)
= 1 - 0.9854 (using NORMSDIST() in excel)
= 0.0146

31.
population mean,u = 6.8
standard deviation,sigma = 1.6

P(X> 7.1 ) = P(Z> ((7.1-6.8) / 1.6)
= P(Z> 0.188 )
= 1 - P(Z<0.188)
= 1 - 0.5746
= 0.4254
= 42.54%

33.
population mean,u = 249.8
standard deviation,sigma = 2.3

P(X>k) = 0.22
P(X<k) = 0.78
P(Z<(k-u)/sigma) = 0.78

(k-u)/sigma = 0.772 [=NORMSINV(0.78) = 0.7722]
k = 0.772*2.3 + 249.8
= 251.6