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Ludmilka [50]
2 years ago
15

By what percentage did the median earnings of college degreed exceed that of high school degreed for 1973 for men (to the neares

t tenth)?___ %
By what percentage did the median earnings of college degreed exceed that of high school degreed for 1967 for women (to the nearest tenth)?___%

Mathematics
1 answer:
kap26 [50]2 years ago
4 0

Remark

I'm going to state what I think the question is saying. I'm having a little trouble figuring it out.

1. Both groups are from 1967.

2. Both groups are women?

It's actually clear, but I had to restate it to get it clear for me.

Solution

<em><u>Givens</u></em>

College: Women= 4400

H.School: Women = 3300

Solve

The base number is the high  school women.

The difference is 4400 - 3300 = 1100

So the % = (1100/3300) * 100% = 33.3%

1973

I did this in the wrong order, but it will still give you the right answer.

The base number is again high school: 5600

Difference: 6600 - 5600 = 1000

% = (1000/5600) * 100% = 17.9

Comment

What this means is that if you take the % difference and multiply it by the smaller number you get the actual difference between the 2 groups.  Put another way, for the 1973 group, the college women made 17.9 % more than the high school group. As an aside, there are lots of reasons for this and I'm not certain it is any longer true.

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Answer:

(a) The standard deviation of the amount spent is $3229.18.

(b) The probability that a household spends between $4000 and $6000 is 0.2283.

(c) The range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.

Step-by-step explanation:

We are given that the average annual amount American households spend on daily transportation is $6312 (Money, August 2001). Assume that the amount spent is normally distributed.

(a) It is stated that 5% of American households spend less than $1000 for daily transportation.

Let X = <u><em>the amount spent on daily transportation</em></u>

The z-score probability distribution for the normal distribution is given by;

                          Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = average annual amount American households spend on daily transportation = $6,312

           \sigma = standard deviation

Now, 5% of American households spend less than $1000 on daily transportation means that;

                      P(X < $1,000) = 0.05

                      P( \frac{X-\mu}{\sigma} < \frac{\$1000-\$6312}{\sigma} ) = 0.05

                      P(Z < \frac{\$1000-\$6312}{\sigma} ) = 0.05

In the z-table, the critical value of z which represents the area of below 5% is given as -1.645, this means;

                           \frac{\$1000-\$6312}{\sigma}=-1.645                

                            \sigma=\frac{-\$5312}{-1.645}  = 3229.18

So, the standard deviation of the amount spent is $3229.18.

(b) The probability that a household spends between $4000 and $6000 is given by = P($4000 < X < $6000)

      P($4000 < X < $6000) = P(X < $6000) - P(X \leq $4000)

 P(X < $6000) = P( \frac{X-\mu}{\sigma} < \frac{\$6000-\$6312}{\$3229.18} ) = P(Z < -0.09) = 1 - P(Z \leq 0.09)

                                                            = 1 - 0.5359 = 0.4641

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                    P(X > x) = 0.03   {where x is the required range}

                    P( \frac{X-\mu}{\sigma} > \frac{x-\$6312}{3229.18} ) = 0.03

                    P(Z > \frac{x-\$6312}{3229.18} ) = 0.03

In the z-table, the critical value of z which represents the area of top 3% is given as 1.88, this means;

                           \frac{x-\$6312}{3229.18}=1.88                

                         {x-\$6312}=1.88\times 3229.18  

                          x = $6312 + 6070.86 = $12382.86

So, the range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.

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