Question

In: Statistics and Probability

The weekly wages of steel workers at a steel plant are normally distributed with a mean...

The weekly wages of steel workers at a steel plant are normally distributed with a mean weekly wage of $940 and a standard deviation of $85. There are 1540 steel workers at this plant. a) What is the probability that a randomly selected steel worker has a weekly wage i) Of more than $850? ii) Between $910 and $960? b) What percentage of steel workers have a weekly wage of at most $820? c) Determine the total number of steel workers with a weekly wage within one standard deviation of the mean. d) Find the lowest and highest weekly wages for the middle 70% of the wage scale.

Solutions

Expert Solution

Given = 940 and = 85

To find the probability, we need to find the z scores.

Since n = 1,

_____________________________________________________

(a) (i) For P (X > 850) = 1 - P (X < 850), as the normal tables give us the left tailed probability only.

For P( X < 850)

Z = (850 – 940)/85 = -1.06

The probability for P(X < 850) from the normal distribution tables is = 0.1446

Therefore the required probability = 1 – 0.1446 = 0.8554

_____________________________________________________

(ii) For P (910 < X < 960) = P(X < 960) – P(X < 910)

For P( X < 960)

Z = (960 – 940)/85 = 0.24

The probability for P(X < 960) from the normal distribution tables is = 0.5948

For P( X < 910)

Z = (910 – 940)/85 = -0.35

The probability for P(X < 910) from the normal distribution tables is = 0.3632

Therefore the required probability is 0.5948 – 0.3632 = 0.2316


_____________________________________________________

(b) For P( X 820) = P(X < 820)

Z = (820 – 940)/85 = -1.41

The required probability from the normal distribution tables is = 0.0793

_____________________________________________________

(c) Within 1 standard deviation means a value below the mean 940 - 85 = 855 and a value above the mean = 940 + 85 = 1025

For P (855 < X < 1025) = P(z = 1) – P(z = -1)

The probability for z = 1 is 0.8413 and for z = -1 is 0.1587

Therefore the required probability is 0.8413 – 0.1587 = 0.6826

_____________________________________________________

(d) The middle 70%

This means that the remaining 30% will be distributed equally to the right and the left = 0.30//2 = 0.15

So the lower p value = 0.15 and the Upper p value = 1 - 0.15 = 0.85

The Z score at p = 0.15 and 0.85 are -1.0364 and +1.0364 respectively.

The Lower value: (X - 940)/85 = -1.0364. Solving for X, X = (-1.0364 * 85) + 940 = $851.9

The Upper value: (X - 940)/85 = +1.0364. Solving for X, X = (+1.0364 * 85) + 940 = $1028.1

_____________________________________________________


Related Solutions

Weekly gross income earned by lecturers in XYZ University is normally distributed with a mean of...
Weekly gross income earned by lecturers in XYZ University is normally distributed with a mean of $1,600 and a standard deviation of $250. The Chief Operation Manager of XYZ University would like to do an audit on the weekly gross income earned by lecturers in his institution. Assist the manager is answering this: what is the probability that if 50 lecturers are randomly selected, their average weekly gross income would be more than $1,700? a. 0.9954 b. 0.9977 c. 0.0023...
Weekly gross income earned by lecturers in XYZ University is normally distributed with a mean of...
Weekly gross income earned by lecturers in XYZ University is normally distributed with a mean of $1,600 and a standard deviation of $250. The Chief Operation Manager of XYZ University would like to do an audit on the weekly gross income earned by lecturers in his institution. Assist the manager is answering this: what is the probability that if 50 lecturers are randomly selected, their average weekly gross income would be more than $1,700? a. 0.9954 b. 0.9977 c. 0.0023...
The annual wages of American farm laborers in 1926 were normally distributed with a mean of...
The annual wages of American farm laborers in 1926 were normally distributed with a mean of $586 and a standard deviation of $97. In 1926, a) What percentage of American farm laborers had an annual wage between $500 and $700? b) What percentage of American farm laborers had an annual wage above $400? c) Find the 75 th Percentile of the farm laborers’ annual wage please be specific
The time it takes workers to finish cleaning is normally distributed with a mean of 73.27...
The time it takes workers to finish cleaning is normally distributed with a mean of 73.27 minutes and a standard deviation of 7.04 minutes. If 13 workers are randomly selected, what is the probability the average time it takes them to finish cleaning is more than 73 minutes?
Depths of pits on a corroded steel surface are normally distributed with mean 820 μm and...
Depths of pits on a corroded steel surface are normally distributed with mean 820 μm and standard deviation 29 μm. a) Find the 10th percentile of pit depths. b) A certain pit is 780 μm deep. What percentile is it on? (Round up the final answer to the nearest whole number.)
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...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT