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velikii [3]
4 years ago
15

In July 2019, there were 1.5 million people who were marginally attached to the labor force, essentially unchanged from a year e

arlier. These are people who were not in the labor force, although they wanted work and were available for it and had looked for jobs at some point in the previous 12 months. They were not counted as unemployed because they had not actively searched for work in the 4 weeks preceding the survey.
SAT
1 answer:
topjm [15]4 years ago
4 0

These people are referred to as the  <u>discouraged workers.</u>A person who is eligible for employment and can work but is unemployed currently and has not attempted to find employment in the last four weeks

Explanation:

Discouraged workers are the individuals who is eligible for employment and can work but is unemployed currently and had not made any attempt to find employment in the last four weeks.

For example: Young graduates,senior citizens.

Discouraged workers are not considered as a part of the labour force and are  not counted in most official unemployment rates.

<u>These are people who were not in the labor force, although they wanted work and were available for it and had looked for jobs at some point in the previous 12 months. They were not counted as unemployed because they had not actively searched for work in the 4 weeks preceding the survey.</u>

If unemployed workers become discouraged, the measured unemployment rate will fall

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Answer;

a) 0.434

b) 0.983

c) 0.367

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The exact question with the given parameters wasn't obtained online, but the same question, albeit with different parameters is then obtained. Hopefully, this Helps to solve the complete question with the required parameters.

Antique Accents tracks their daily profits and has found that the distribution of profis is approximately normal with a mean of $17,700.00 and a standard deviation of about $900.00. Using this information, answer the following questions For full marks your answer should be accurate to at least three decimal places. Compute the probability that tomorrow's profit will be

a) less than $16,791 or greater than $18,231

b) greater than $15,783

c) between $17,997 and $20,130

Solution

This is a normal distribution problem with

Mean = μ = $17,700

Standard deviation = σ = $900

a) less than $16,791 or greater than $18,231. P(x < 16,791) or P(X > 18,231) = P(X < 16,791) + P(x > 18,231)

We first standardize 16,791 and 18,231

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 16791

z = (x - μ)/σ = (16791 - 17700)/900 = - 1.01

For 18231

z = (x - μ)/σ = (18231 - 17700)/900 = 0.59

To determine the required probability

P(X < 16,791) + P(x > 18,231) = P(z < -1.01) + P(z > 0.59)

We'll use data from the normal probability table for these probabilities

P(X < 16,791) + P(x > 18,231) = P(z < -1.01) + P(z > 0.59)

P(z < -1.01) = 0.15625

P(z > 0.59) = 1 - (z ≤ 0.59) = 1 - 0.7224 = 0.2776

P(X < 16,791) + P(x > 18,231) = P(z < -1.01) + P(z > 0.59) = 0.15625 + 0.2776 = 0.43385 = 0.434 to 3 d.p

b) greater than $15,783. P(x > 15783)

We standardize 15783

z = (x - μ)/σ = (15783 - 17700)/900 = -2.13

To determine the required probability

P(x > 15783) = P(z > -2.13)

We'll use data from the normal probability table for this probability

P(x > 15783) = P(z > -2.13) = 1 - P(z ≤ - 2.13)

= 1 - 0.01659 = 0.98341 = 0.983 to 3 d.p.

c) between $17,997 and $20,130.

P(17,997 < x < 20,130)

We first standardize 17,997 and 20,130

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 17,997

z = (x - μ)/σ = (17,997 - 17700)/900 = 0.33

For 20,130

z = (x - μ)/σ = (20,130 - 17700)/900 = 2.70

To determine the required probability

P(17,997 < x < 20,130) = P(0.33 < x < 2.70)

We'll use data from the normal probability table for these probabilities

P(17,997 < x < 20,130) = P(0.33 < x < 2.70)

= P(z < 2.70) - P(z < 0.33)

= 0.99653 - 0.62930

= 0.36723 = 0.367 to 3 d.p.

Hope this Helps!!!

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