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# Probability, Sampling Distributions, and Inference

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Probability, Sampling Distribuons, and Inference
Probability, Sampling Distributions, and Inference
Chapter 5
1) The working of 1, 2 and 3 standard deviations (σ) with a mean (µ) of 2500 hours is given
below and as graphically plotted there under.
-1σ to +1 σ is between 2000 hours and 3000 hours of light bulbs
-2σ to +2 σ is between 1500 hours and 3500 hours of light bulbs
-3σ to +3 σ is between 1000 hours and 4000 hours of light bulbs

2
Probability, Sampling Distribuons, and Inference
We can observe from the above graph that the distribution is symmetric with a mean of 2500
hours, therefore, 50% of light bulbs will have a life of less than 2500 hours and the other 50% of
light bulbs will have a life of more than 2500 hours.
2) The standard deviation (σ) is 5 hours and the mean (µ) is 370 hours. We can solve this
question by using the graph as above or it can be solved by using standard score (z) formula and
its table. We will be using standard score (z) formula and table to solve the problem in order to
get the precise percentage. The working of 1 standard deviation (σ) with a mean (µ) of 370 hours
is as follows.
-1σ to +1 σ is between 365 hours (370 – 5) and 375hours (370 + 5) of light bulbs
Standard score (z) = [(Data value – Mean) / Standard deviation]. Here we are going to find two
standard scores i.e. 365 hours and 375 hours to find out the percentage on either side of 1
standard deviation.
Standard score (z) of 365 hours will be [(365 – 370) / 5] = -1. Now we can refer the table of
standard score (z) as given on page 211 of the course book (Bennett, Briggs & Triola, 2009). So,
the percentage corresponding to standard score of -1 is 15.87%.
Now we are going to calculate standard score of 375 hours i.e. (375-370) / 5 = 1. The
corresponding percentage to the standard score of +1 is 84.13%.
The difference between those percentages will be the percentage of bulbs that have lifetimes
lying within 1 standard deviation of the mean on either side. Therefore, the percentage difference
between 84.13% and 15.87% (84.13 – 15.87) is 68.26%.

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Probability, Sampling Distribuons, and Inference
3) The mean (µ) of the phone bill is \$60 and standard deviation (σ) is \$12. Now we have to
calculate 68% of the phone bill which is 1 standard deviation (σ). The calculation of 1 standard
deviation (σ) which is 68% will be as follows.
-1σ will be (\$60 – \$12) = \$48
+1σ will be (\$60 + \$12) = \$72
Therefore, 68% of Jen’s phone bills are between \$48 and \$72.
4) The 25
th
percentile standard score (z) is approximately -0.65 according to standard score
and percentile table given on page 211 of the course book (Bennett, Briggs & Triola, 2009).
When we multiply this standard score (z) of -0.65 to standard deviation of \$10 (-0.65 x\$10), we
get - \$6.50, i.e. 25
th
percentile is \$6.50 below the mean (µ) of \$50 phone bill. Therefore, we will
deduct \$6.50 from mean (µ) of \$50 phone bill (\$50 - \$6.50) = \$43.50 which is 25
th
percentile.
(Note: This is an approximate value because the standard score table in the course book is
limited to one digit value after the decimal point).
5) Firstly, we are going to find the standard score (z) with a data value of 0.32 inches to get
the corresponding percentage. The following is the formula of standard score (z).
Standard score = [(Data value - Mean) / Standard deviation]
Standard score = [(0.32 – 0.30) / .01] = 2.0
The corresponding percentage of standard score (z) 2.0 is 97.72%. To find the percentage of
bolts with a diameter greater than 0.32 inches, we have to deduct 97.72% from 100% to get the
answer, therefore, (100% - 97.72%) = 2.28% of bolts will have a diameter greater than 0.32
inches.

### Unformatted Attachment Preview

Probability, Sampling Distributions, and InferenceChapter 51)The working of 1, 2 and 3 standard deviations (?) with a mean () of 2500 hours is given below and as graphically plotted there under.-1? to +1 ? is between 2000 hours and 3000 hours of light bulbs -2? to +2 ? is between 1500 hours and 3500 hours of light bulbs-3? to +3 ? is between 1000 hours and 4000 hours of light bulbsWe can observe from the above graph that the distribution is symmetric with a mean of 2500 hours, therefore, 50% of light bulbs will have a life of less than 2500 hours and the other 50% of light bulbs will have a life of more than 2500 hours.2)The standard deviation (?) is 5 hours and the mean () is 370 hours. We can solve this question by using the graph as above or it can be solved by using standard score (z) formula and its table. We will be using standard score (z) formula and table to solve the problem in order to get the precise percentage. The working of 1 standard deviation (?) with a mean () of 370 hours is as follows. -1? to +1 ? is between 365 hours (370 - 5) and 375hours (370 + 5) of light bulbsStandard score (z) = [(Data value - Mean) / Standard deviation]. Here we are going to find two standard scores i.e. 365 hours and 375 hours to find out the percentage on either side of 1 standard deviation. Standard score (z) of 365 hours will be [(365 - 370) / 5] = -1. Now we can refer the table of standard score (z) as given on page 211 of the course book (Bennett, Briggs & Tri ...
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