Normal Distribution Flashcards
1
Q
What is a Distribution?
A
The shape of a data set that has been plotted as a histogram.
2
Q
Which are examples of data plotted as a histogram?
A
- Uniform = all same.
- Multimodal = 2 peaks.
- Bernoulli = 50-50.
- Unimodal = 1 peak.
- Skewed = large numbers in end.
3
Q
How is the histogram plotted?
A
y axis = No of observations.
x axis = Measure.
4
Q
What do measurements follow?
A
Normal distribution.
5
Q
What is the peak in a normal distribution curve?
A
Mean = Mode = Median.
6
Q
What are the characteristics of the normal distribution curve?
A
Bell - shaped curve.
Symmetrical.
7
Q
What is a small sample we can plot in a normal distribution and how do they look?
A
Heights in our age group.
Roughly symmetrical.
8
Q
What is a medium sample we can use in a normal distribution curve?
A
All 1st year students.
9
Q
What is a large sample we can use in a normal distribution curve?
A
Population of Dundee.
10
Q
What is a very large sample we can use in a normal distribution curve?
A
Population of Scotland.
11
Q
What can we measure to give us a normal distribution curve?
A
Entire population of earth.
12
Q
Where will 95% of values in a population lie in the graph?
A
Between 1.96 standard deviations of the mean (mean +/- 1.96sd).
13
Q
What can all normal distributions have?
A
Different shapes.
14
Q
What is the relationship between a sample and a mean estimation?
A
Larger sample –> better mean estimation.
15
Q
What is μ and σ?
A
μ = true population mean. σ = true population standard deviation.
16
Q
What is x and s?
A
x = sample mean. sd = sample standard deviation.
17
Q
Why are measurements normally distributed?
A
Determined by many sources of variation.
18
Q
Why is height normally distributed?
A
People have different environments.
Height is determined by genes.
Small errors occur in measurements.
19
Q
What are the 2 factors that influence height?
A
50% individuals have increased height by 20%.
50% individuals have lowered height by 20%.
20
Q
What do we understand by the 2 factors that influence height?
A
Half of individuals have height = 1205 of average.
Half of individuals have height - 80% of average.
21
Q
How can we form these 2 factors that influence height in a different percentage?
A
Increased height by 10%
Lowered height by 10%.
22
Q
How many possible combinations do we have if we have 10% increased/lowered height as influencing factors?
A
4.
23
Q
What are the 4 possible combinations of the 10% factors?
A
- short - long = 100% = 1/4 prevalence.
- long - short = 100% = 1/4 prevalence.
- short-short = 80% = 1/4 prevalence.
- long - long = % = 1/4 prevalence.
24
Q
How will the graph be plotted based on the 10% increased/lowered height factors?
A
25% = 80% individuals. 50% = 100% individuals. 25% = 120% individuals.
25
Which can be the factors for length?
Increased height by 5%.
| Lowered height by 5%.
26
How many will the possible combinations be for the 5% factors?
16.
27
What are the percentages of the 5% factors?
```
80.
90.
100.
110.
120.
```
28
How many possible combinations can be found based on 1 factor?
2.
80%.
120%.
29
How many possible combinations can be found based on 2 factors?
3.
80%.
100%.
120%.
30
How many possible combinations can be found based on 4 factors?
```
5.
80%.
90%.
100%.
110%.
120%.
```
31
How many possible combinations can be found based on 8 factors?
8.
32
What is different in normal distribution curves?
Means.
| Widths.
33
What are the 2 parameters that describe a normal distribution?
Mean = μ.
| Standard deviation = σ.
34
What do we say if a variable X follows a normal distribution?
X is N(μ, σ2).
| Χ-Ν((μ, σ2).
35
Determine mean chest measurement and standard deviation if:
| The chest measurements (variable X) in inches* of 5732 Scottish soldiers is normally distributed as X~N(40,4).
X =inches.
| sd = inches.
36
What is inches?
cm.
| For length.
37
Compete description of normal distribution for dataset to 2 decimal places if:
At the turn of the 19th /20th centuries, data were collected on the length of criminals’ left middle fingers. The data were normally distributed with mean length 11.55cm and standard deviation 0.55cm.
X-N(11.55, 0.30).
38
Determine mean nicotine level in sample and sd if:
The blood plasma nicotine levels (X) in ng/ml were determined in 55 smokers and found to follow the normal distribution model below.
X~N(324, 21567)
```
Mean = 324.
Sd = 147.
```
39
What are the features of normal distribution?
Typical bell shaped curve.
Smoothed histogram.
Unimodal.
40
By what is the width of the distribution described?
Standard deviation = σ.
41
Where do 68% of all measurements lie?
Within 1 standard deviation of the mean.
42
Where do 95% of all measurements lie?
Within 1.96 standard deviations of the mean.
43
What do normal distributions with different parameters have?
Same % of data.
44
What is the general result for the normal distribution?
In a normally distributed population, proportion of measurements lying within z standard deviations of the mean is the same regardless of parameters μ and σ.
45
What is the normal distribution if z=1?
68%.
46
What is the normal distribution if z = 1.96?
95%.
47
How can the area under any portion of a normal distribution be solved?
By using a single standard distribution curve.
48
Where do we use One-tailed plots?
To find one side of mean.
| 95% observations and 5% observations.
49
Where do we use a Two-tailed plot?
To find either sides of mean.
| 95% observations and 2.5% and 2.5% observations.
50
What do we select to find two-tailed plot?
0 up to Z.
51
What does this equation mean:
| μ +/- 0,37σ.
z = 0.37/
Need +/- 0.37 --> 2 x 14.43.
Move bubble until z = 0.37 in the curve.
52
What does the equation mean:
| μ + 0,80σ.
z = 0.80.
Need +0.80 --> 1 x 28.81.
Move bubble in curve until z = 0.80.
53
What happens in a normal distribution table?
Z to 1 decimal place = vertically.
x to 2nd decimal place = horizontally.
Find number x 100 --> percentage.
54
What is rarely feasible?
To determine population mean and standard deviation.
55
How do we estimate μ and σ?
By taking samples of appropriate size.
56
How do determine x and s?
By estimating μ and σ respectively.
57
What should sample be?
Population representative.
58
What will the mean heart rate for each group of 10 patients be?
Variable between 67-83bpm.
59
How can we make the mean heart rate settle at 75bpm?
With cumulative mean.
| Increase sample size from 10 to 120.
60
What will the standard deviation of the mean heart rate for each group of 10 be?
Variable between 6.9-19.3 bpm.
61
How can we make the standard deviation of mean heart rate settle at 14bpm?
Increase sample size from 10 to 120.
| Cumulative sd of heart rate.
62
What is Central Limit Theorem about?
How variable our estimate of population mean is.
63
What happens as sample size increases?
Better estimates of μ and σ.
| Decreased x and s fluctuations.
64
What are the sample means?
Population estimates.
65
With what are all estimates associated?
Error.
66
What will the sample mean have?
Distribution.
67
Why will the sample mean have distribution?
Because samples of patients had different mean heart rates.
| Random variation.
68
When can we achieve a normal distribution curve?
When groups size increases.
Have distribution in group mean.
= Central Limit theorem.
69
What happens even if the original population is not normal?
Sampling distribution.
| Gets normal as group sample size increases.
70
Why is Central Limit Theorem important?
In laboratory we use small samples with normally distributed means.
Analyse data using statistical methods for normal distributed data.
71
What is standard about?
How variable our estimate of population mean is.
72
What does standard error of mean quantify?
How much on average sample means differ from population mean.
73
What can we do with standard error as we can not know it with certainty?
Estimate it.
74
How can we calculate standard error estimation?
SE = s / square root of N.
75
What is the relationship between sample and standard error based on their equation?
Larger sample --> smaller standard error.
76
What does it mean in statistics when a symbol has a bar across top?
It is an estimate of population value --> derived from a sample.
77
What does standard error provide?
Measure of estimating uncertainty.
78
What is N?
How many samples we have in total.
79
What will the number of samples estimate?
Variability in population activity.
80
What do the 3 times the samples are measured tell us?
Pseudo replicates.
If pipetting is good.
Take mean.
81
What do series of experimental measurements/observations provide?
Sample mean.
| Standard deviation.
82
What should we do not confuse?
Standard error.
| Standard deviation.
83
What shall we use when we calculate standard error?
Correct N value.
84
What does the Central Limit Theorem allow us to assume?
Means of small samples are normally distributed.
Use powerful statistical tests for analysis.
85
What is the sample mean?
An estimate of population mean.
86
What do we get every time we measure?
Different results.
87
What does the 95% confidence interval provide?
A range of plausible values for the population mean based on our sample data.
88
What can we be then based on 95% confidence interval?
95% confident true population mean lies in this interval.
89
What is μ-?
Lower limit.
| Boundary of confidence interval.
90
What is μ+?
Upper limit.
91
How is the plot if a population is normally distributed?
μ- = lower limit --> probability to be greater than mean is 0.025/2.5%.
x = sample mean.
μ+ = upper limit --> probability of being smaller than mean is 0.025/2.5%.
92
What are the values for the upper 95% confidence limit μ+ for population?
```
x = μ+ - 1.96 σ/square root of n.
μ+ = x + 1.96 σ/square root of n/
```
93
How are the intervals of the 95% confidence intervals for population of lower confidence limit?
```
x = μ- + 1.96 σsquare root of n.
μ- = x - 1,96 σ/square root of n.
```
94
When does the formulae of upper limit and lower limit work?
When the population standard deviation is known.
When we have a very large sample. n > 80.
95
What happens if the formulae of upper and lower limit does not work?
σ cannot be estimated accurately.
| Need alternative for small samples.
96
Why do we need an alternative method for small samples?
Because when we measure activity for 5 experiments 2 times each s is not reliable.
97
Why is s not reliable for small samples?
It varies from 0 to 4.03.
98
What would have been better for a properly designed experiment?
N = 3.
99
What is the equation of finding confidence limits for a sample?
(μ+, μ-) = x +/- 1.96 σ / square root of n.
100
With what can we replace σ / square root of n?
SE = s / square root of N.
101
What is SE?
An estimate.
102
What happens when we replace 'σ / square root of n' with SE = s / square root of N?
Sample mean will not vary.
Normal distribution.
Replace Z = 1.96 with a value from a different distribution = t-distribution.
103
What do means of large samples and populations have?
A normal distribution.
104
What is a single small dataset for a population?
n<50.
| Indefinite estimate.
105
What do the means of smaller samples have based on Gossett?
1. Similar, flatter, wider distribution compared to normal.
2. Flatters as sample size decreases.
3. 'degrees of freedom' shows how flat they are.
106
What happens to the t-distribution as n increases?
Becomes identical with normal distribution.
107
What does each curve represent in a normal distribution plot?
Each degree of freedom.
108
On what does the shape of curve depend?
Degrees of freedom.
109
How is the quantity (n-1) known if there are n observations in a sample?
Degrees of freedom of sample.
110
How many observations have value if we know the mean?
2.
111
```
By using the medcalc website we have to find the t value when the CI=95% on the column 0.95, 0.05 for:
degrees of freedom = 3.
sample size (n) = 8.
degrees of freedom = 25.
degrees of freedom >500.
```
3. 182
2. 306
2. 060
1. 960
112
What is the N value for large sample in normal distribution?
N = Z.
113
What do we have to modify to find the confidence limits for a sample:
(μ-, μ+) = x +/- 1.96 σ/square root of n.
| replace σ/square root of n with SE = s/ square root of N.
114
What is the SE?
An estimate.
115
What happens to the sample x due to the fact that SE is an estimate?
It does not vary precisely.
| Have to be replaced Z = 1.96 with a value from a different distribution = t-distribution.
116
Which equation we have to use to find the confidence limits for a sample?
(CL+, CL-) = x +/- tdf SE.
Find t-tables when df = n-1.
SE = s/ square root of N.
117
For which samples is the equation of confidence limits valid?
For all sample sizes with normal distribution.
118
From where is the sample mean x and standard deviation in the equation of confidence limits?
From experimental observation.
119
The mean urinary lead concentration in 140 children was 2.18 mol/24 h, with standard deviation 0.87.
What is the standard error of the mean to 3 dec. places?
SE = s/square root of N
0. 87/square root of 140
0. 074.
120
The mean urinary lead concentration in 140 children was 2.18 mol/24 h, with standard deviation 0.87.
Calculate the 95% confidence intervals to 2 dec. places
(CL+, CL-) = x +/- tdfSE
t139 = 1.978
2.18+/-1.978 x 0.074
(CL-, CL+) = 2.03, 2.33
121
What can be carried out in medicine?
Measurements on blood sample to test for disease.
122
What can be the interest of a clinician?
A biomarker for a disease.
123
Which are some examples of biomarkers for a disease?
High/low hormone concentrations.
High/low enzyme activity.
High/low concentration of a substance in the blood.
124
What happens if the biomarker for a disease is very high/very low?
Problem.
125
What will the biomedical scientists in the lab compare?
Patient's measurement to a reference range based on 95% CI.
126
Why will the biomedical scientists compare patient's measurements to a reference?
To assist diagnosis.
127
Which are some examples of reference range in medicine?
Ferritin = 12-30 ng/mL(men).
= 12-150 ng/mL (women).
Glucose = 65-110 mg/dL.
128
What happens if men have ferritin levels below 12 ng/mL?
Anaemia.
129
What happens to men when exceed 300 ng/mL of ferritin?
Hemochromatosis.
130
How is the reference range worked out in medicine?
With very large samples.
| Determining values where 95% of population is μ+/-1,96σ.
131
What happens if patient's value is outside/near top/bottom of reference range?
Health problem.
132
cholesterol levels in mg/dL (X) for women ages 20-34 are normally distributed as X~N(185,1521).
Q: What are the chances that a woman in this age group would have a cholesterol level greater than 240 mg/dL?
Would this level be a cause for concern?
```
μ = 185, σ = 39 square root of 1521.
x = 240
(240-185) = 55 mg/dL above mean value.
```
133
cholesterol levels in mg/dL (X) for women ages 20-34 are normally distributed as X~N(185,1521).
```
Q: What are the chances that a woman in this age group would have a cholesterol level greater than 240 mg/dL?
Would this level be a cause for concern?
How many (z) standard deviations (s) above the mean (m) is 240?
```
Z = 55/square root of 1521 = 1.41
| Patient has a cholesterol level that is 1.41 standard deviations above the mean.
134
Using the normal distribution tool online, determine probability of patient with high level of cholesterol.
One tailed = 'up to Z'
92.07%.
Patient's cholesterol levels is in highest 8% of population and would be a cause of concern.
135
A break in has occurred and crime scene investigators have found a chisel mark on the frame of a window. The width if the mark was 14.0mm. A suspect who was carrying a chisel was arrested nearby and test marks were made on samples of wood similar to the window frame using the suspect’s chisel. The following were results obtained.
Width of chisel mark (mm)14.5, 14.1, 13.8, 14.4, 14.2, 14.7, 14.2, 14.4, 14.0, 14.2, 14.2, 14.7, 14.2, 13.9, 14.6, 14.2
Is it plausible that the mark on the window frame was made by the suspect’s chisel i.e. does the chisel mark from the crime scene lie within the 95% confidence interval for the marks made with the suspect’s chisel?
```
Find confidence limits;
(CL+, CL-) = x +/- tdf SE.
(CL+, CL-) = x +/- tdf s/square root of N.
Mean width of 16 marks = 14.27mm.
Standard deviation = 0.265mm.
Normally distributed.
SE = 0.265/square root of 16 =
0.265/4= 0.0663.
tdf = t(16-1) = t15 = 2013
(CL+, CL-) = 14.27 +/- (2.13 x 0.0663)=
(14.13, 14.41)
95% confidence interval is (14.1, 14.4).
```
136
Is there evidence to implicate the suspect?
Mark of window = 14.0mm
Analysis indicated 95% CI (14.1, 14.4)
95% certain true mean width of mark left by suspect lies between 14.1mm and 14.4mm.
Mark on window was 14.0mm
lies outside interval
95% confident mark was not made by suspect.
137
The mean fat content of the 13 biscuits was 16.60g with a standard deviation of 2.066g. The data were normally distributed.
Calculate the 95% confidence interval for the sample. Should Miriam accept the claim that her biscuits have on average 15g of fat each?
```
(CL+, CL-) = x +- tdf SE
SE = s/square root of N.
SE = 2.066/square root of 13
SE = 2.066/3.6056 = 0.573.
tdf = t(13-1) = t12 = 2.18
(CL+, CL-) = 16.60 +- (2.18x0.573) = (15.351, 17.849).
```
The 95% confidence interval is (15.35, 17.85).
138
The manufacturer claimed that the biscuits contained 15g of fat
Miriam’s analysis indicated that the 95% confidence interval was (15.35,17.85)
Miria is 95% certain that true value of mean fat content lies between 15.35g and 17.85g.
139
Is the manufacturer's claim correct?
Manufacturer's value of 15g lies outside Miriam's interval.
Miriam is 85% confident fat content exceeds manufacturer's claimed value.
140
What type of variable is weight of patient (kg)?
Continuous.
