Question

In: Statistics and Probability

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 accepted.

Solutions

Expert Solution

Solution :

Given that ,

mean = = 195.2

standard deviation = = 0.8

n = 22

m = 195.2

m =  / n = 0.8 / 22=0.1706

P(M <195.1 ) = P[( - m ) / m < (195.1-195.2) / 0.1706]

= P(z <-0.586 )

Using z table  

= 0.2789

probability=0.2789   


Related Solutions

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 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!
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 106.4-cm and a standard deviation of 1.8-cm. For shipment, 9 steel rods are bundled together. Round all answers to four decimal places if necessary. A. What is the distribution of X? X ~ N(     ,     ) B. What is the distribution of ¯x? ¯x ~ N(      ,    ) C. For a single randomly selected steel rod, find the probability that the length...
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 207.2-cm and a standard deviation of 2.3-cm. For shipment, 17 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 207.4-cm and 207.9-cm. For 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 127.3-cm and a standard deviation of 1.7-cm. Find the proportion of steel rods with lengths between 130.7 cm and 132.2 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.
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 226.9-cm and a standard deviation of 1.6-cm. For shipment, 47 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 227-cm and 227.1-cm. For 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 93.6-cm and a standard deviation of 1.6-cm. For shipment, 30 steel rods are bundled together. Find P32, which is the average length separating the smallest 32% bundles from the largest 68% bundles. P32 =___________ -cm Enter your answer as a number accurate to 2 decimal place.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT