TASK 4 Sold Price’ sorted in order from lowest to highest. 112 114 152 230 239.5
ID: 3439992 • Letter: T
Question
TASK 4
Sold Price’ sorted in order from lowest to highest.
112
114
152
230
239.5
295
305
310
323
340
342
347
350
357
358.5
375
375
410
432.5
445
446
455
460
465
475
481
490
491
510
525
552
560
563
596
615
618
670
700
711
882.5
951
1060
1175
1225
1360
1885
2020
TASK 5
SOLD PRICE ($000)
Mean
577.7447
Standard Error
58.47
Median
465
Mode
375
Standard Deviation
400.85
Sample Variance
160677.80
Kurtosis
6.91
Skewness
1.97
Range
1908
Minimum
112
Maximum
2020
Sum
27154
Count
47
QUESTION 1
Calculate, using Standard Normal tables, approximately how many “Sold Price” data values in your sample you would expect to lie within 1.5 standard deviations of the mean (i.e. between z = –1.5 and z = +1.5).
QUESTION 2
Use your sorted “Sold Price” sample data from Task 4, and the mean and standard deviation from the Descriptive Statistics table of Task 5, to manually count the number of “Sold Price” data values in your sample that lie within 1.5 standard deviations of the mean. And State whether this count matches, approximately, your answer to question 1
112
114
152
230
239.5
295
305
310
323
340
342
347
350
357
358.5
375
375
410
432.5
445
446
455
460
465
475
481
490
491
510
525
552
560
563
596
615
618
670
700
711
882.5
951
1060
1175
1225
1360
1885
2020
Explanation / Answer
1.
Using normal tables/technology, the left tailed areas of z = 1.5 and z = -1.5 are
P(z<-1.5) = 0.066807201
P(z<1.5) = 0.933192799
Thus, the area between these two z scores is
P(-1.5<z<1.5) = 0.866385597 [answer]
Thus, there are approximately 0.866385597*47 = 40.72012308
or 41 items within 1.5 standard deviations.
*******************
2.
Here,
Mean = 577.7447
Standard deviation = 400.85
Thus,
Lower bound = Mean - 1.5*standard deviation = -23.5303
Upper bound = Mean + 1.5*standard deviation = 1179.0197
There are 43 items out of 47 that lie between these two numbers. [answer]
This is quite close to the result in Part 1.
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