1 A company produces steel rods The lengths of the steel rod
1. A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 111.1-cm and a standard deviation of 2-cm. Suppose a rod is chosen at random from all the rods produced by the company. There is a 61% probability that the rod is longer than: .....
Answer as a number accurate to 1 decimal place
B. A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 118.1-cm and a standard deviation of 0.5-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 less than 118.2 cm.
Answer as a number accurate to 1 decimal place
C. For a Normal distribution with mean, =2, and standard deviation, =4 what proportion of observations take values less than 0 or greater than 2?
Round to 4 decimal places.
2. The daily milk production of a herd of cows is assumed to be Normally distributed with a mean of 40 liters, and standard deviation of 11.9 liters.
A) On what proportion of days is daily production less than 10.8 liters?
Answer= (Round your answer to 3 decimal places.)
B) On what proportion of days is production more than 10.6 liters?
Answer= (Round your answer to 3 decimal places.)
3. The lengths of pregnancies in a small rural village are Normally distributed with a mean of 266 days and a standard deviation of 13 days.
What percentage of pregnancies last beyond 291 days?
P(X > 291 days) = .... %.
4. A company knows that replacement times for the DVD players it produces are Normally distributed with a mean of 8.3 years and a standard deviation of 1.9 years.
A. Find the proportion of a randomly selected DVD players that will have a replacement time less than 3.9 years?
P(X < 3.9 years) = .....
Answer to 4 decimal places.
B. If the company wants to provide a warranty so that only 2.6% of the DVD players will be replaced before the warranty expires, what is the time length of the warranty?
warranty = ..... years
Solution
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2) the probability that the length of this rod is less than 118.2 cm.
= P(Z<3.6)
c)
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2) As z value = -7.4, prob = 0
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| mu | 111.1 | |
| sigma | 2 | |
| z | -1.23 | |
| x | 108.64 | 108.6 |

