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Vladimir79 [104]
3 years ago
15

0.25 + 1 + 4 + 16 + 64 Enter the values used in finding a partial sum.

Mathematics
2 answers:
Deffense [45]3 years ago
6 0
This is a geometric series

r = 4
a1 = 0.25
n = 5
Alexeev081 [22]3 years ago
4 0

Answer:  The required values are

r=4,~~a_1=0.25~~\textup{and}~~n=5.

Step-by-step explanation:  We are given a partial sum as follows :

0.25 + 1 + 4 + 16 + 64.

We are to find the values of r,~~a_1~~\textup{and}~~n.

From the given partial series, we note that

the series is a geometric one with first term 0.25 and common ratio calculated as follows :

r=\dfrac{1}{0.25}=\dfrac{4}{1}=\dfrac{16}{4}=\dfrac{64}{16}=4.

Also, since there are 5 terms in the given sum, so the value of n is 5.

Thus, the required values are

r=4,~~a_1=0.25~~\textup{and}~~n=5.

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Compute the mean of the data as follows:

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The median for odd set of values is the computed using the formula:

Median=(\frac{n+1}{2})^{th}\ obs.

Arrange the data set in ascending order as follows:

36, 38, 43, 44, 44, 45, 50, 55, 56, 56, 57, 60, 62, 65, 69

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(b)

The first quartile is the middle value of the upper-half of the data set.

The upper-half of the data set is:

36, 38, 43, 44, 44, 45, 50

The middle value of the data set is 44.

Thus, the first quartile is 44.

The third quartile is the middle value of the lower-half of the data set.

The upper-half of the data set is:

56, 56, 57, 60, 62, 65, 69

The middle value of the data set is 60.

Thus, the third quartile is 60.

(c)

The range of a data set is the difference between the maximum and minimum value.

Maximum = 69

Minimum = 36

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Range =Maximum-Minimum\\=69-36\\=33

Thus, the value of range is 33.

The inter-quartile range is the difference between the first and third quartile value.

Compute the value of IQR as follows:

IQR=Q_{3}-Q_{1}\\=60-44\\=16

Thus, the inter-quartile range is 16.

(d)

Compute the variance of the data set as follows:

s^{2}=\frac{1}{n-1}\sum (x_{i}-\bar x)^{2}\\=\frac{1}{15-1}[(55-52)^{2}+(56-52)^{2}+...+(65-52)^{2}]\\=100.143

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Q_{1}-1.5QR=44-1.5\times 16=20

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Q_{3}+1.5QR=60-1.5\times 16=80

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Thus, there are no outliers in the data set.

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