Question

In: Statistics and Probability

A company produces steel chains with an average breaking strength of ?=200 lbs with ?=10 lbs....

A company produces steel chains with an average breaking strength of ?=200 lbs with ?=10 lbs.


Suppose you take a sample of n=100 chains. What is the probability that the sample mean breaking strength will be 201 lbs or more?

0.4413
0.8587
nearly 0
0.1587

Solutions

Expert Solution

Let denotes the sample mean breaking strength for  a sample of n=100 chains.

or

Now,

The probability that the sample mean breaking strength will be 201 lbs or more

orchestra answered 3 years ago
Next > < Previous

Related Solutions

In a sample of 75 steel wires, the average breaking strength is 50 kN, with a...
In a sample of 75 steel wires, the average breaking strength is 50 kN, with a standard deviation of 1.9 kN. a) Find a 99% confidence interval for the mean breaking strength of this type of wire. b) An engineer claims that the mean breaking strength is between 49.7 kN and 50.3 kN With what level of confidence can this statement be made? c) How many wires must be sampled so that a 99% confidence interval specifies the mean breaking...
3. A company manufactures rope whose breaking strengths have a mean of 300 lbs with a...
3. A company manufactures rope whose breaking strengths have a mean of 300 lbs with a standard deviation of 24 lbs. It is believed that a new process will increase the mean breaking strength of the ropes. a) Design a decision rule for rejecting the old process at the = 0.01 level if a sample of 64 ropes is chosen. b) Under the decision rule of part (a), what is the probability of accepting the old process when in fact...
Suppose you believe the strength of the linear association between weight (lbs) and sleep (average per...
Suppose you believe the strength of the linear association between weight (lbs) and sleep (average per night), r1=-0.2 is significantly different than the association between weight and exercise (hours per week), r2=-.25. What is the required sample size required to assess whether the two correlations are significantly different? Assume alpha = 0.05 and beta = 0.1.    
A company produces wooden chairs and tables. Each chair uses 10 lbs. of wood and takes...
A company produces wooden chairs and tables. Each chair uses 10 lbs. of wood and takes 10 hours to construct. Each table uses 15 lbs. of wood and takes 5 hours to construct. There are up to 150 lbs. of wood available and up to 110 labor hours available. If the profit on each chair is $3 and the profit on each table is 4$. We wish to maximize the profit. Let x be the number of chairs produced and...
A. A company produces steel rods. The lengths of the steel rods are normally distributed with...
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...
A company produces steel rods. The lengths of the steel rods are normally distributed with a...
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 167.1-cm and a standard deviation of 0.6-cm. For shipment, 6 steel rods are bundled together. Round all answers to four decimal places if necessary. What is the distribution of XX? XX ~ N( , ) What is the distribution of ¯xx¯? ¯xx¯ ~ N( , ) For a single randomly selected steel rod, find the probability that the length is between 166.9-cm...
A company produces steel rods. The lengths of the steel rods are normally distributed with a...
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 195.2-cm and a standard deviation of 0.8-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 less than 195.1-cm. P(M < 195.1-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...
A company produces steel rods. The lengths of the steel rods are normally distributed with a...
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 143.6-cm and a standard deviation of 0.8-cm. For shipment, 41 steel rods are bundled together. Round all answers to four decimal places if necessary. What is the distribution of X ? X ~ N(,) What is the distribution of ¯x ? ¯x ~ N(,) For a single randomly selected steel rod, find the probability that the length is between 143.4-cm and 143.5-cm....
A company produces steel rods. The lengths of the steel rods are normally distributed with a...
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 222-cm and a standard deviation of 1.5-cm. For shipment, 9 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is less than 222.1-cm. P(M < 222.1-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...
A company produces steel rods. The lengths of the steel rods are normally distributed with a...
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 124.7-cm and a standard deviation of 1.6-cm. For shipment, 17 steel rods are bundled together. Find P83, which is the average length separating the smallest 83% bundles from the largest 17% bundles. P83 = -cm Please provide a step-by-step!
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT