Use the information in
[link] , but change the sample size to 144.
Find
P (20<
<30).
Find
P (
Σx is at least 3,000).
Find the 75
th percentile for the sample mean excess time of 144 customers.
Find the 85
th percentile for the sum of 144 excess times used by customers.
Solutions
0.8623
0.7377
23.2
3,441.6
In the United States, someone is sexually assaulted every two minutes, on average, according to a number of studies. Suppose the standard deviation is 0.5 minutes and the sample size is 100.
Find the median, the first quartile, and the third quartile for the sample mean time of sexual assaults in the United States.
Find the median, the first quartile, and the third quartile for the sum of sample times of sexual assaults in the United States.
Find the probability that a sexual assault occurs on the average between 1.75 and 1.85 minutes.
Find the value that is two standard deviations above the sample mean.
Find the
IQR for the sum of the sample times.
We have,
μ
x =
μ = 2 and
σ
x =
=
= 0.05. Therefore:
50
th percentile =
μ
x =
μ = 2
25
th percentile = 1.97
75
th percentile = 2.03
We have
μ
Σx =
n (
μ
x ) = 100(2) = 200 and
σ
μx =
(
σ
x ) = 10(0.5) = 5. Therefore
50
th percentile =
μ
Σx =
n (
μ
x ) = 100(2) = 200
25
th percentile = 196.63
75
th percentile = 203.37
P (1.75<
<1.85) = 0.0013
Using the
z -score equation,
, and solving for
x , we have
x = 2(0.05) + 2 = 2.1
The
IQR is 75
th percentile – 25
th percentile = 203.37 – 196.63 = 6.74
Try it
Based on data from the National Health Survey, women between the ages of 18 and 24 have an average systolic blood pressures (in mm Hg) of 114.8 with a standard deviation of 13.1. Systolic blood pressure for women between the ages of 18 to 24 follow a normal distribution.
If one woman from this population is randomly selected, find the probability that her systolic blood pressure is greater than 120.
If 40 women from this population are randomly selected, find the probability that their mean systolic blood pressure is greater than 120.
If the sample were four women between the ages of 18 to 24 and we did not know the original distribution, could the central limit theorem be used?
P (
x >120) = 0.0272. There is about a 3%, that the randomly selected woman will have systolics blood pressure greater than 120.
P (
>120) = 0.006. There is only a 0.6% chance that the average systolic blood pressure for the randomly selected group is greater than 120.
The central limit theorem could not be used if the sample size were four and we did not know the original distribution was normal. The sample size would be too small.
A study was done about violence against prostitutes and the symptoms of the posttraumatic stress that they developed. The age range of the prostitutes was 14 to 61. The mean age was 30.9 years with a standard deviation of nine years.
In a sample of 25 prostitutes, what is the probability that the mean age of the prostitutes is less than 35?
Is it likely that the mean age of the sample group could be more than 50 years? Interpret the results.
In a sample of 49 prostitutes, what is the probability that the sum of the ages is no less than 1,600?
Is it likely that the sum of the ages of the 49 prostitutes is at most 1,595? Interpret the results.
Find the 95
th percentile for the sample mean age of 65 prostitutes. Interpret the results.
Find the 90
th percentile for the sum of the ages of 65 prostitutes. Interpret the results.
P (
<35) = 0.9886
P (
>50) ≈ 0. For this sample group, it is almost impossible for the group’s average age to be more than 50. However, it is still possible for an individual in this group to have an age greater than 50.
P (
Σx ≥ 1,600) = 0.0864
P (
Σx ≤ 1,595) = = 0.9005. This means that there is a 90% chance that the sum of the ages for the sample group
n = 49 is at most 1595.
The 95th percentile = 32.7. This indicates that 95% of the prostitutes in the sample of 65 are younger than 32.7 years, on average.
The 90th percentile = 2101.5. This indicates that 90% of the prostitutes in the sample of 65 have a sum of ages less than 2,101.5 years.
Bacteria doesn't produce energy they are dependent upon their substrate in case of lack of nutrients they are able to make spores which helps them to sustain in harsh environments
_Adnan
But not all bacteria make spores, l mean Eukaryotic cells have Mitochondria which acts as powerhouse for them, since bacteria don't have it, what is the substitution for it?
Assimilatory nitrate reduction is a process that occurs in some microorganisms, such as bacteria and archaea, in which nitrate (NO3-) is reduced to nitrite (NO2-), and then further reduced to ammonia (NH3).
Elkana
This process is called assimilatory nitrate reduction because the nitrogen that is produced is incorporated in the cells of microorganisms where it can be used in the synthesis of amino acids and other nitrogen products
There are nothing like emergency disease but there are some common medical emergency which can occur simultaneously like Bleeding,heart attack,Breathing difficulties,severe pain heart stock.Hope you will get my point .Have a nice day ❣️
_Adnan
define infection ,prevention and control
Innocent
I think infection prevention and control is the avoidance of all things we do that gives out break of infections and promotion of health practices that promote life
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Source:
OpenStax, Statistics i - math1020 - red river college - version 2015 revision a - draft 2015-10-24. OpenStax CNX. Oct 24, 2015 Download for free at http://legacy.cnx.org/content/col11891/1.8
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