14 inch document....it is reduced to 0.6 times as long...
14 * 0.6 = 8.4 inches long
it is too small, so he enlarges it 1.4 times...
8.4 * 1.4 = 11.76 inches <===
Answer:
The following are the answer to this question:
Step-by-step explanation:
In the given question the numeric value is missing which is defined in the attached file please fine it.
Calculating the probability of the distribution for x:
The formula for calculating the mean value:
use formula for calculating the Variance:
![\to \bold{\text{Variance}= E(X^2) -[E(X)]^2}](https://tex.z-dn.net/?f=%5Cto%20%5Cbold%7B%5Ctext%7BVariance%7D%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%7D)
calculating the value of standard deivation:
Standard Deivation (SD) = 
(a) Data with the eight day's measurement.
Raw data: [60,58,64,64,68,50,57,82],
Sorted data: [50,57,58,60,64,64,68,82]
Sample size = 8 (even)
mean = 62.875
median = (60+64)/2 = 62
1st quartile = (57+58)/2 = 57.5
3rd quartile = (64+68)/2 = 66
IQR = 66 - 57.5 = 8.5
(b) Data without the eight day's measurement.
Raw data: [60,58,64,64,68,50,57]
Sorted data: [50,57,58,60,64,64,68]
Sample size = 7 (odd)
mean = 60.143
median = 60
1st quartile = 57
3rd quartile = 64
IQR = 64 -57 = 7
Answers:
1. The average is the same with or without the 8th day's data. FALSE
2. The median is the same with or without the 8th day's data. FALSE
3. The IQR decreases when the 8th day is included. FALSE
4. The IQR increases when the 8th day is included. TRUE
5. The median is higher when the 8th day is included. TRUE
Answer:
A ) i) X control chart : upper limit = 50.475, lower limit = 49.825
ii) R control chart : upper limit = 1.191, lower limit = 0
Step-by-step explanation:
A) Finding the control limits
grand sample mean = 1253.75 / 25 = 50.15
mean range = 14.08 / 25 = 0.5632
Based on X control CHART
The upper control limit ( UCL ) =
grand sample mean + A2* mean range ) = 50.15 + 0.577(0.5632) = 50.475
The lower control limit (LCL)=
grand sample mean - A2 * mean range = 50.15 - 0.577(0.5632) = 49.825
Based on R control charts
The upper limit = D4 * mean range = 2.114 * 0.5632 = 1.191
The lower control limit = D3 * mean range = 0 * 0.5632 = 0
B) estimate the process mean and standard deviation
estimated process mean = 50.15 = grand sample mean
standard deviation = mean range / d2 = 0.5632 / 2.326 = 0.2421
note d2 is obtained from control table
Answer: The lowest value: 100 and highest value: - 150 .
If we were to build the box plot for this data, the box would stretch between 
and 
.
Step-by-step explanation:
We know that the box-plot is the graphical way to represent the five -number summary (Minimum value , First quartile 
, Median , Third Quartile 
, Maximum value).
Where , the box streches between the first quartile and the third quartile .
Given : The cost of taking your pet aboard the air flight with you in the continental US varies according to the airlines.
The five number summary for prices based on a sample of major US airlines was:
Min = 60,
Median = 110
Max = 150
If we were to build the box plot for this data, the box would stretch between and .
Hence, the lowest value: 100 and highest value: 125 .
