Question #189702

The following table shows the age distribution of live births in Albany County, New York, for 2000.​

Mother’s Age

Number of Deaths due to Accidents

10-14

7

15-19

258

20-24

585

25-29

841

30-34

981

35-39

526

40-44

99

45-49

4

(a) For the data Calculate Mean, median and Mode.

(b) Which average is most appropriate and why?

(c) Comment on the Shape of the distribution from (a).



1
Expert's answer
2021-05-07T14:12:41-0400

Solution:


Mean=ΣfixiΣfi=952473301=28.85Mean=\dfrac{\Sigma f_ix_i}{\Sigma f_i}=\dfrac{95247}{3301}=28.85

Next, N/2 = 3301/2 = 1650.5

So, median class is 24.5-29.5

Then, l=24.5,h=5,cf=850,f=841l=24.5, h=5,cf=850,f=841

Median=l+(N2cff)×h=24.5+(1650.5850841)×5=29.26Median=l+(\dfrac{\frac N2-cf}{f})\times h=24.5+(\dfrac{1650.5-850}{841})\times 5=29.26

Next, modal class is 29.5-34.5

f1=981,f0=841,f2=526,l=29.5f_1=981,f_0=841,f_2=526,l=29.5

So, mode=l+(f1f02f1f0f2)×h=29.5+(9818412(981)841526)×5mode=l+(\dfrac{f_1-f_0}{2f_1-f_0-f_2})\times h=29.5+(\dfrac{981-841}{2(981)-841-526})\times 5

=30.67=30.67

(b) Mode seems most appropriate as biggest frequencies (i.e., 841,981) lies around 30.

(c) Curve goes steep up around mid-classes and goes steep down at the end, which shows the data is mostly spread around mid-values/mid-classes.


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