Inferiential statistics .pptx

UNIT – V INFERENTIAL STATISTICS
1
Mrs. D. Melba Sahaya Sweety RN,RM
PhD Nursing , MSc Nursing (Pediatric Nursing),BSc
Nursing
Associate Professor
Department of Pediatric Nursing
Enam Nursing College, Savar, Bangladesh.

NORMAL DISTRIBUTION
INTRODUCTION
• Normal distribution, also known as the Gaussian distribution, is a probability
distribution that is symmetric about the mean, showing that data near the mean are
more frequent in occurrence than data far from the mean.
• In graphical form, the normal distribution appears as a "bell curve".
• In a normal distribution the mean is zero and the standard deviation is 1. It has zero
skew and a kurtosis of 3.
• Normal distributions are symmetrical, but not all symmetrical distributions are
normal. Symmetrical distributions occur when where a dividing line produces two
mirror images.
• Many naturally-occurring phenomena tend to approximate the normal distribution.
2

NORMAL DISTRIBUTION
• The normal distribution curve is bell-shaped.
• The mean, median, and mode are equal and located at the
center of the distribution.
• The normal distribution curve is unimodal (single mode).
• In a normal distribution curve, the greatest proportions of score
lie close to the mean.
• Almost all the scores (0.9997 of them) lie within the SD of the
mean ( Mean ± 3 SD = 99.73).
• It is asymptotic, i.e the distribution graph approached towards
end but never touch the horizontal line.
• The total area under the normal distribution curve is equal to
1or 100 % 3
PROPERTIES OF NORMAL DISTRIBUTION

NORMAL DISTRIBUTION
• It is Continuous for all values of X between -∞ and ∞
• The notation N(µ, σ 2 ) means normally distributed with mean
µ and variance σ 2 . If we say X ∼ N(µ, σ 2 ) we mean that X is
normally distributed with mean µ, and Variance σ2.
• Two parameters, µ and σ. Note that the normal distribution is
actually a family of distributions, since µ and σ determine the
shape of the distribution. The mean μ tells about location -
Increase μ - Location shifts right ,Decrease μ – Location shifts
left Shape is unchanged The variance σ2 tells about narrowness
or flatness of the bell - Increase σ2 - Bell flattens. Extreme
values are more likely Decrease σ2 - Bell narrows. Extreme
values are less likely Location is unchanged
4

NORMAL DISTRIBUTION
5
Curve Position or shape (relative
to standard normal
distribution)
A (M = 0, SD = 1) Standard normal
distribution
B (M = 0, SD = 0.5) Squeezed, because SD < 1
C (M = 0, SD = 2) Stretched, because SD > 1
D (M = 1, SD = 1) Shifted right, because M > 0
E (M = –1, SD = 1) Shifted left, because M < 0

NORMAL DISTRIBUTION
FORMULA
The probability density function of normal or Gaussian distribution is given by;
6
x = variable
X: -∞ to +∞
e = Euler’s constant = 2.71828
π = mathematical constant = 3.14 note: π = (circumference/diameter) for any circle.
μ = population mean
σ = Standard Deviation

NORMAL DISTRIBUTION
• The area under the normal curve that lies within
 68% of data falls within the first standard deviation from the mean.
 95% of the values fall within two standard deviations from the mean..
 99.7% of data will fall within three standard deviations from the mean.
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EMPRICAL RULE OR THREE SIGMA RULE
OR 65-95-99.7 RULE

NORMAL DISTRIBUTION
8
Example : 1 The normal distribution below has a standard
deviation of 10. Approximately what area is contained between 70
and 90 ?
10 20 30 40 50 60 70 80 90 100 110 120
47.5%
µ = 70
σ = 10

NORMAL DISTRIBUTION
9
Example : 2 The normal distribution below, Approximately what
area is contained between - 2 and 1 ?
-3 -2 -1 0 1 2 3
µ = 0
σ = 1
47.5% 34%
81. 5%

CENTRAL LIMIT THEOREM
 The central limit theorem states that the sampling distribution of the mean approaches a
normal distribution, as the sample size increases. This fact holds especially true for sample
sizes over 30. As the sample size increases, the distribution of frequencies approximates a
bell-shaped curved (i.e. normal distribution curve).
 Therefore, as a sample size increases, the sample mean and standard deviation will be closer
in value to the population mean μ and standard deviation σ . A sufficiently large sample can
predict the parameters of a population such as the mean and standard deviation.
This is useful, as the research never knows which mean in the sampling distribution is the
same as the population mean, but by selecting many random samples from a population the
sample means will cluster together, allowing the research to make a very good estimate of the
population mean. 10

STANDARD NORMAL DISTRIBUTION
OR Z - DISTRIBUTION
• The standard normal distribution, also called the z-
distribution, is a special normal distribution where the mean is
0 and the standard deviation is 1.
• Any normal distribution can be standardized by converting its
values into z scores. Z scores tell how many standard deviations
from the mean each value lies.
• The formula to transform every normal random variable X into
a z score is z = (X – μ) / σ
• A “z” –value is the distance between a selected value , designated X,
where X is a normal random variable or individual value, μ is
the mean of x , and σ is the standard deviation of x .
11

1, To the left of any Z value
12
OR Z - DISTRIBUTION
EMPRICAL RULE OF Z DISTRIBUTION
Finding Areas Under the Standard Normal Distribution Curve :
P(z<-a) = P(z<a) = Q(a)
-a 0 0 a
f(z) f(z)
=

2, To the Right of any Z value
13
STANDARD NORMAL
DISTRIBUTION OR Z - DISTRIBUTION
P(z>-a) = P(z>a) = 1 - Q(a)
-a 0
f(z) f(z)
0 a
=

3, Between any two Z values
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STANDARD NORMAL
DISTRIBUTION OR Z - DISTRIBUTION
P(-b<z<-a) = P(a<z<b) =
P(z>a) – P(z>b) = Q(a) – Q(b)
-b -a 0
f(z) f(z)
0 a b
=

Example 1, Find the area of the left Z = 1.89
15
OR Z - DISTRIBUTION
P(z< 1.89) = 0.97062
0 1.89
f(z)

16
OR Z - DISTRIBUTION
Example 2, Find the Probability of Z greater than 2.0
Area
Required
Z = 0 Z=2.0
P(Z > 2.0) = 1 – P(Z >2.0)
P(Z >2.0 ) = 1- 0.9772
P(Z >2.0 ) = 0.0228

17
OR Z - DISTRIBUTION
Example 3, Find the area under the curve P(-2.0 ≤ z ≤ 0.95)
Area
Required
Z = -2.0 Z = 0 Z = 0.95
P(-2.0 ≤ z ≤ 0.95) = P (Z< 0.95) - P (Z < -2.0)
P(-2.0 ≤ z ≤ 0.95) = 0.8289 - 0.0228
P(-2.0 ≤ z ≤ 0.95) = 0.8061

18
OR Z - DISTRIBUTION
Example 4, Find the area under the curve P(-1.5 ≤ z ≤ -0.5)
Area
Required
Z = -1.5 ,Z= -0.5 ,Z = 0
P(-1.5 ≤ z ≤ -0.5) = P (Z< -0.5) - P (Z < -1.5)
P(-1.5 ≤ z ≤ -0.5) = 0.3085 – 0.0668
P(-1.5 ≤ z ≤ -0.5) = 0.2417

19
OR Z - DISTRIBUTION
Example 5, Find Z if 81.9% of the area is to the right of z
Area
Required
Z= -0.91 ,Z = 0
P ( z > 0.8190) = 1 – 0.8190
= 0.1810
= - 0.9

20
OR Z - DISTRIBUTION
Example 6, Find the P(x = 16)
µ= 0 µ= 16
P(X = 16) = 0

21
OR Z - DISTRIBUTION
Example 7, Find the P(X< 54) with the mean of 65 and
standard deviation 9
x - µ
σ
Z =
Z = 54 –65
9
= - 11/9
Z = - 1.22
µ= 54 µ= 65
Area
Required
Z = -1.22 Z = 0
P(X < 54 ) = P (Z< -1.22)
= 0.1112

22
OR Z - DISTRIBUTION
Example 9, Find the P(x ≥ 80) with the mean of 65 and
x - µ
σ
Z =
Z = 80 – 65
9
= 15 / 9
Z = 1.67
µ= 65 µ= 80
Area
Required
Z = 0 Z = 1.67
P(X > 80) = P (Z > 1.67)
= 1 - P (Z < 1.67)
= 1 - 0.9525
P(X > 80) = 0.0475

23
OR Z - DISTRIBUTION
Example 10, Find the P(x > 20) with the mean of 12 and
x - µ
σ
Z =
Z = 20 – 12
4
= 8/4
Z = 2
µ= 12 µ= 20
Area Required
Z = 0 Z = 2
P(X >20) = P (Z>2)
= 1– 0.97725
P(X >20) = 0.0228

24
OR Z - DISTRIBUTION
Example 11, Find the Probability of x between 70 and 86
with the mean of 65 and standard deviation 9
x - µ
σ
Z =
Z = 70 – 65
9
= 5 / 9
Z = 0.56
µ= 65,µ= 70,µ= 80
Area
Required
Z = 0,Z = 0.56,Z = 2.33
P(70 < X< 86) = P (0.56 < Z< 2.33)
= P (Z < 2.33) – P (Z < 0.56)
= 0.9901 – 0.7123
P(70 <X > 86) = 0.2778
Z = 86 – 65
9
= 21 / 9
Z = 2.33

25
OR Z - DISTRIBUTION
Example 12, Find the P(x ≤ 8) with the mean of 12 and
x - µ
σ
Z =
Z = 8 –12
4
= - 4/4
Z = - 1
µ= 8 µ= 12
Area
Required
Z = -1 Z = 0
P(X < 8 ) = P (Z< - 1 )
P(X < 8) = 0.1587

26
OR Z - DISTRIBUTION
Example 13 , Find the Probability of P ( X≤16)
P ( X ≥ 12)
P ( X≤16)
P ( X≤16) = P ( Z<1)=0.8413
P ( X ≥ 12) = 1 – P (Z > 0)
= 1- 0.5000 = 0.5
P ( X≤16) = 0.8413
P ( X ≥ 12) 0.5
P ( X≤16)
P ( X ≥ 12) = 0.6826
µ= 12 µ= 16
Z = 0 Z = 1
P ( X ≥ 12)
P (Z =0 ∩ Z = 1)

27
OR Z - DISTRIBUTION
Example 14, Find the Z – Score corresponding to the 20th Percentile
Z = -0.84 Z = 0
20th Percentile = 0.20
Z = -0.84

28
OR Z - DISTRIBUTION
Example 15, Find what score represents the 85th percentile with µ = 20 and σ = 3
Z = 0 Z = 1.04
85 th Percentile = 0.85
Z = 1.04
X = 20 + (1.04 x 3)
X = 20 + 3.12
X = 23.12
Z =
x - µ
σ
Z σ = X - µ
µ + Z σ =X
X = µ + Z σ 0.85

29
OR Z - DISTRIBUTION
Example 16, Find 8 % of score are above what score with µ = 20 and σ = 3
Z = 0 Z = 1.41
8% = 0.08
Above 8 % = 1- 0.08 = 0.92
Z = 1.41
X = 20 + (1.41 x 3)
X = 20 + 4.23
X = 24.23
So only 24.23% of scores
are above 8%
Z =
x - µ
σ
Z σ = X - µ
µ + Z σ =X
X = µ + Z σ
0.08

30
OR Z - DISTRIBUTION
Example 17, A 100 watt light bulb has an average brightness of
1640 lumens with a standard deviation of 62 lumens,
A, What is the probability that a 100 watt light bulb will have a
brightness more than 1800 lumens
x - µ x = 1800 , µ = 1640 , σ = 62
σ
Z =
Z = 1800 – 1640
62
= 160/62
Z = 2.58
Area
Required
Z = 0 Z = 2.58
P(X>1800) = P (Z> 2.58)
= 1 – 0.9951
= 0.049

31
OR Z - DISTRIBUTION
Example 17, B, What is the probability that a 100 watt light
bulb will have a brightness less than 1550 lumens
x - µ x = 1550 , µ = 1640 , σ = 62
σ
Z =
Z = 1550 – 1640
62
= -90/62
Z = -1.452
Area
Required
Z = -1.45 Z = 0
P(X<1550) = P (Z< -1.45)
= 0.0735

32
OR Z - DISTRIBUTION
Example 17, c, What is the probability that a 100 watt light bulb will
have a brightness between 1600 and 1700 lumens
x - µ x = 1600 and 1700 , µ = 1640 , σ = 62
σ
Z =
Z = 1600 – 1640
62
= -40/62
Z = -0.65
Z = 1700 – 1640
62
Z = 60/62
Z = 0.97
Area
Required
Z = -1.45 Z = 0 z = 0.96
P(1600<X<1700) = P (-0.65<Z< 0.96)
= P(Z<0.97) – P(Z<-0.65)
= 0.8340 – 0.2578
= 0.5762

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