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Pavel [41]
3 years ago
5

0.5, 3/16, 0.75, 5/48 least to greatest

Mathematics
1 answer:
jarptica [38.1K]3 years ago
7 0
An easy way for me to do this is to either make all of the fractions into decimals or all the decimals into fractions, whatever you prefer. I prefer fractions, but for typing reasons I’m going to make them all decimals. Making 5/48 into a decimal you divide 5 by 48 and get 0.104... and divide 3 by 16 and get 0.1875. Then from there, look at whatever tenths place number is the least (the number after the decimal). Two of them have one, so moving onto the next number, 8>0, which means 0.104 is less then 0.1875. Then 5<7. So from least to greatest, 5/48 (0.104), 3/16 (0.1875), 0.5, 0.75.
Good Luck!
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Answer:

a) Mean= 15, standard deviation= 5.1575

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c) See below

Step-by-step explanation:

(a) Compute the mean and standard deviation for the 10 cases for which x is not missing.

The mean using the ten known values is

\bar x=\frac{17+6+12+14+20+23+9+12+16+21}{10}=15

The standard deviation is

\small s=\sqrt{\frac{(17-15)^2+(6-15)^2+(12-15)^2+(14-15)^2+(20-15)^2+(23-15)^2+(9-15)^2+(12-15)^2+(16-15)^2+(21-15)^2}{10}}=5.1575

(b) Create a new data set with 20 cases by using imputation where you set the values for the 10 missing cases equal to the mean that you computed in the previous part. Compute the mean and standard deviation for this new data set with 20 cases.

The new data set would be

17, 6, 12, 14, 20, 23, 9, 12, 16, 21, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15

The new mean for this set of values would be then

\bar x=\frac{\displaystyle\sum_{i=1}^{20}x_i}{20}=15

The new standard deviation is now

s=\sqrt{\frac{\displaystyle\sum_{i=1}^{20}(x_i-\bar x)^2}{20}}=3.6469

(c) Summarize what you have learned about the possible effects of this type of imputation on the mean and the standard deviation.

Obviously, this way of imputation is somewhat arbitrary and will produce a set of data undoubtedly skewed.  

It could possibly be a sensible way of imputing data if the amount of missing data is very little compared to the whole set, for example one or two data in a set of 100, but in this case we have 50% of missing data and it makes no sense this procedure.

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