Question

1 point) The following stem-and-leaf plot the number of items produced per day in a factory in a sample of 24 randomly selected days 112 represents 120 Leaf Unit 10.0 7 0 2 3 6 7 7 8 8 5 9 9 0 1 5 6 103 6 (a) What is the value of the mode for this sample? 630, 650, 77(if there is more than one answer, enter them separated by commas. If there is no answer, type the word none in the box.) (b) What is the value of the median? 745 (c) What is the range of this data set? 680 (d) The mean of this sample is 757.5 and the standard deviation is 162.6. If the statement at the beginning of the stem-and leaf plot was 12 represents 1.20°,what would be the value of the mean of the data represented in that stem and leaf plot? 7.575 the standard deviation of that data? 1.626 (e) According to Chebyshev's rule, how many observations should lie within one and a half standard deviations of the mean? 13 Using the mean and standard deviation given in (d), how many observations actually lie within one and a half standard deviations of the mean? 24

I just keep getting the last two wrong.

Answer 1

Solution:

Now, we have to find out the:

d) What would be the mean of the data represented in the stream and standard deviation:

Given ,

the mean of the sample = 757.5

standard deviation = 162.6

• According to the documentation, the esteem of the mean of the information spoke to in that stem and leaf plot is 7.57.5
• And the estimation of the standard deviation of the information spoke to in that stem and leaf plot is 1.626.

e) Now, we have to find out the:

How many observations should lie with one and half standard deviation of the mean, According to chebyshev rule :

We know the formula of chebyshev's rule is $\bar{x} \pm ks$

For lower bound:

$\bar{x} - ks$ =  757.5 - k ( 162.6 )

Here k = one and half standard deviation

k= 1 1/2

$\bar{x} - ks$ = 757.5 - 1 1/2 ( 162.6 )

= 757.5 -1.5( 162.6 )

= 757.5 - 243.9

= 513.6

$\bar{x} - ks$=  513.6

$\therefore$ For lower bound the one and half standard deviation of the mean is = 513.6

For higher bound:

$\bar{x} + ks$ ,

$\bar{x} + ks$ = 757.5 + 1 1/2 ( 162.6 )

= 757.5 +1.5( 162.6 )

= 757.5 + 243.9

$\bar{x} + ks$   = 1001.4

$\therefore$ For higher boundary the one and half standard deviation of the mean is = 1001.4

Number of observations actually lie within one and half standard deviation of mean :

Number of observations actually lie with in lower and upper bound .

i.e., Number of observations = ( 513.6 & 1001.4 )

In the question, From the data we can see the no.of observation lie with in the boundaries range.

From the notation , lower bound 513.6 = 5.136

and  lower bound 1001.4 = 10.014

so, the no.of observations boundaries range is 5.136 to 10.014

From the data , we can get the observation is " 5.6,6.0,6.1,6.3,6.3,6.5, 6.5, 6.9 7.0,7.2,7.3,7.6,7.7,7.7,7.8,8.5,8.9,9.0,9.5, 9.6 "

Total number of observations = 21

$\therefore$ Number of observations actually lie within one and half standard deviation of mean =21