Question

In: Statistics and Probability

The life expectancy of a person in 24 randomly selected countries for the year 2011 is...

  1. The life expectancy of a person in 24 randomly selected countries for the year 2011 is in the table below.

    a.) What is the range of the life expectancy rates? b.) What is the median of the life expectancy rates?

    77.2

    55.4

    69.9

    76.4

    75.0

    78.2

    73.0

    70.8

    82.6

    68.9

    81.0

    54.2

    5) Cholesterol levels were collected from patients two days after they had a heart attack and are shown in the table below. (Show all work. Just the answer, without supporting work, will receive no credit.)

    270

    236

    210

    142

    280

    272

    160

    220

    226

    242

    186

    266

    206

    318

    294

    282

    234

    224

    276

    282

    360

    310

    280

    278

    288

    288

    244

    236

    1. a.) What is the sample mean?

    2. b.) What is the sample standard deviation? (Round your answer to two decimal

      places.

    3. 6) There are 4 black marbles and 6 red marbles in a box. Consider selecting one marble at a time from the box. What is the probability that the first marble is black and the second marble is also black. Express the probability in fraction format. (Show all work. Just the answer, without supporting work, will receive no credit.) a.) Assume the marble is selected with replacement. b.) Assume the marble is selected without replacement.

Solutions

Expert Solution

( 4 )

Range :

Ordering the data from least to greatest, we get:

54.2   55.4   68.9   69.9   70.8   73.0   75.0   76.4   77.2   78.2   81.0   82.6   

The lowest value is 54.2.

The highest value is 82.6.

The range = 82.6 - 54.2 = 28.4.

The range of data set is 28.4.

Median :

The median is the middle number in a sorted list of numbers. So, to find the median, we need to place the numbers in value order and find the middle number.

Ordering the data from least to greatest, we get:

54.2   55.4   68.9   69.9   70.8   73.0   75.0   76.4   77.2   78.2   81.0   82.6   

As you can see, we do not have just one middle number but we have a pair of middle numbers, so the median is the average of these two numbers:

Median = ( 73.0+75.0) / 2 = 74

The median of the data set is 74.

( 5 )

( a )  sample mean = Sum of terms / Number of terms

= 7110 / 14

= 253.9286

( b )  sample standard deviation :

Create the following table.

data data-mean (data - mean)2
270 16.0714 258.28989796
236 -17.9286 321.43469796
210 -43.9286 1929.72189796
142 -111.9286 12528.01149796
280 26.0714 679.71789796
272 18.0714 326.57549796
160 -93.9286 8822.58189796
220 -33.9286 1151.14989796
226 -27.9286 780.00669796
242 -11.9286 142.29149796
186 -67.9286 4614.29469796
266 12.0714 145.71869796
206 -47.9286 2297.15069796
318 64.0714 4105.14429796
294 40.0714 1605.71709796
282 28.0714 788.00349796
234 -19.9286 397.14909796
224 -29.9286 895.72109796
276 22.0714 487.14669796
282 28.0714 788.00349796
360 106.0714 11251.14189796
310 56.0714 3144.00189796
280 26.0714 679.71789796
278 24.0714 579.43229796
288 34.0714 1160.86029796
288 34.0714 1160.86029796
244 -9.9286 98.57709796
236 -17.9286 321.43469796

∑(xi−X bar )2=61459.8571

S = sqrt( 61459.8571 / 27 )

= 47.7105

S = 47.71


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