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Normal distribution, inverse normal, hypothesis testing, finding mean and SD, converting binomial to normal.

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uNmPHypnhsfHaSafrBQE

Study note

Ahmed Nour ✓™

United States of America

DEFAULT (US)

DEFAULT

Statistics

Normal Distributional

Normal distribution

Normal distribution calculations

NORMAL DISTRIBUTION
N = mean
0² = vanance
20
o
99.71.
20
mean = mode = median
68% of data lies between one standard deviation of mean
95%
2 standard deviations of mean
3 standard deviations of Mean
"
Example 1
A
7.28 8.2
(standard deviation ²)
example
X~N (30,42)
symmetrical
mean
X~ N(8,0.2)
normal
6 The weights of a group of dormice,
D grams, are modelled as D~ N(, 25).
If 97.5% of dormice weigh less than
70 grams, find u.
Upper-33
5 The lengths of a colony of adders, Ycm, are modelled as Y~ N(100, 2). If 68% of the adders
have a length between 93 cm and 107 cm, find o².
68% Jone standard deviation
107-93=142 = 7
a) P(X<33)
lower mean - so
= 30-20-10
A
3033
a) P(x >8)
= 0.5
vanance b) P(7.8 < X <8.2)
FINDING PROBABILITIES
normal cumulative distribution (normal CD)
..
X<33
= 0.68
Standard deviation = 5
70=p+ 2x5
70= N +10
N=60
0²=7²=49
97.5% less than 70, the 2.5% greater than 10.
↑
into calc. X-0.77
book 2 chapter 3.
one o each side
100-95-5-2
2 standard deviations
6) P(x ≥24) = P(X>24)
= 0.9332
c) P (33.52X<38.2) = 0.1706
30 d) P(X<27 or X >32)
27 30 32
1- P(27<x<32)
3) Y~ N(25,25)
a) P(Y220)
= 1-0.46 0.535
Ex 3 B
1) X~ N(30,2²)
a) P(x<33) b) P(X>26) C) P(x>31.6)
= 0.933 ✓
= 0.977 ✓
Yo s
9 = 0.159 ✓
5) M~N(15,1.52)
a) i) P (M> 14)
b) P (1824 226)
= 0.499✓
8) X~N (6,0.8²)
a) i) P(X77)
= 0.7475 V
b) sum of answers = 1
ii) P(M< 14)
= 0.1056 ✓
b) P(x 25) = 0.1056
(c) P(X25)
= 0.1056 ✓
Ans³= 1.179 y 10-3✓
lower
umit
to find a in P(X<a).
1- prev. Ans= 0.252 ✓
as all probabilities add up to 1/
=
INVERSE NORMAL DISTRIBUTION
c) P(Y>23.8)
= 0.595 ✓
11 The diameters of bolts, D mm, made by a particular machine are modelled as D~N(13,0.1²).
a Find the probability that a bolt, chosen at random, has a diameter less than 12.8 mm.
(1 mark)
a) P(X<12.8) = 0.0227 b) P(12.9 4X413.1)
lower = 0.
upper = 12.8
O = 0.1
N = 15
upper
umit
P(X731.6)
= 0.2118 ✓
Bolts are considered to be 'perfect' if the diameter lies between 12.9 mm and 13.1 mm.
A random sample of 40 bolts is taken.
b Find the probability that more than 25 of the bolts are 'perfect'.
on calc.
X~ N(20, 32). Find, correct to two decimal places, the values of a such that:
a P(X<a) = 0.75
area
b P(X> a) = 0.4
c P(16 < X <a) = 0.3
=
(4 marks)
X~B(40,0.68)
P(X>25)
= 1- P(x≤25) = 0.722 ✓
0.68
probability one is perfect a) a= 22.023
b) P(x > a) = 0.4 • P(x <a) = 0.6
a= 20.76
4 The random variable Y~ N(100, 15¹).
a Find the value of a and the value of b such that:
i P(Y> a) = 0.975
ii P(Y<b) = 0.10
b Find P(a < Y<b).
a) i) P(Y>a)= 0.975
P(yza)= 0.025
a = 70.6
o
to
253-50 =
4
= (0.75)
STANDARD NORMAL DISTRIBUTION
→ mean is O → Standard deviation I
-X~ N(N₂0²) = you can STANDERDIZE
2 = X-N
Finding N
X~N(N₁3²)
Standard
O
ii) P(Y26)=0.1
The random variable X~ N(50, 42). Write in terms of Þ(z) for some value z:
a P(X<53)
b P(X= 55)
a) 2= X-P
flw
will be distributed mean 0 S.D I
4
0.2
6=80.78
c) P(16< X < a) = 0.3
Z
↑ inverse
area=0.2
U=O
=1-(1.25)
0=1
16 20
b) Plazy2b)
b) P(X55) = |- P(x455)
2=55-50 = 5
4
4
a
P(x720)=0.2 find N
↓
P(Z > 20-N)=0.2
3
P(70.64 y 80.78) = 0.075
2=0.8416
P(x<a)= P(X<16) +0.3
= 0.0921 +0.3
P(x/a) = 0.3921
a=19.178
2~N(0,1²)
O
P(zza)
0-8416=20-0
P=17.5
a
need to use STANDARD
NORMAL
= Da
fancccy symbol
2 = x-P Finding o
X~N (50,0²)
0.2119
Z
inverse to find
2 = -0.7998
11 An automated pottery wheel is used to make bowls. The diameter of the bowls, D mm,
is normally distributed with mean and standard deviation 5 mm. Given that 75% of
bowls are greater than 200 mm in diameter, find:
a the value of μ
b P(204<D<206)
N
N = np
Three bowls are chosen at random.
e Find the probability that all of the bowls are greater than 205 mm in diameter.
P(D<200)=0₂²₂2-
P(Z < 200-P)=0.25
0.75
a) P(x246)=0.2119, find o
P(Z < 46-50)=0.2119
-0
Binomial to Normal
X~B(nip)
if nis large
-0.7998 = 46-50
o
and
0 = 46-50 = 5
-0.7998
o = √rip (1-P)
S
inverse normal.
area=0.25 N=00=1
2 =-0.6744
P
is close to 0.5
(2 marks)
(1 mark)
(3 marks)
-0-6747= 200-N
5
-3.37= 200-P
.37 = N₂₁
= 203mm/
you can convert to normal. Hypothesis Testing.
look at MEAN of a sample from population.
X~N(N₁02)
A certain company sells fruit juice in cartons. The amount of juice in a carton has a normal
distribution with a standard deviation of 3 ml.
Mo:
N = 60
H₁ : N260
Lone
tailed
sample
mean
The company claims that the mean amount of juice per carton, u, is 60 ml. A trading inspector has
received complaints that the company is overstating the mean amount of juice per carton and he
wishes to investigate this complaint. The trading inspector takes a random sample of 16 cartons
and finds that the mean amount of juice per carton is 59.1 ml.
in
Using a 5% level of significance, and stating your hypotheses clearly, test whether or not there is
evidence to uphold this complaint.
X~ N(60, 3²)
16
X~N (N₁0²)
n
assume Ho
NC60,0.732)
sample
size
Ho: N=0.580
H₁ N70.580
X~ N(60, 3²)
1
L
I
(
59.1 60
>
standadize
normal CD
like usual
5
PCX 259.1)= 0.1151
0.1151 0.05
So there is not enough.
evidence to reject to, and conclude
the mean amount of juice is less.
than 60ml
relate it
back to question -
If you need to find a
you can standardize you test statistic
X~ N(N₁0²)
CRITICAL value (region
Z = X-P
S
A machine produces bolts of diameter D where D has a normal distribution with mean 0.580 cm
and standard deviation 0.015 cm.
The machine is serviced and after the service a random sample of 50 bolts from the next production
run is taken to see if the mean diameter of the bolts has changed from 0.580 cm. The distribution of
the diameters of bolts after the service is still normal with a standard deviation of 0.015 cm.
a Find, at the 1% level, the critical region for this test, stating your hypotheses clearly.
The mean diameter of the sample of 50 bolts is calculated to be 0.587 cm.
b Comment on this observation in light of the critical region.
D~N (0.58,0.015²)
D~N(0.58,0.015²) 2 failed
chnical regions
at two ends
in verse normal
arca: 0.005
5:1
P: 0
decode data
-2.5758=d-0.58
we have to find a critical region so we standadise
2= D-P
= D-0.58
品
0.015
√50
a) Ho: № = 100
H₁: N2100
2=X-100
X~N(100,15²)
15
√80
= -2.5758
0.015
√50
4 The IQ scores of a population are normally distributed with a mean of 100 and a standard
deviation of 15. A psychologist wishes to test the theory that eating chocolate before sitting an
IQ test improves your score. A random sample of 80 people are selected and they are each given
an identical bar of chocolate to eat before taking an IQ test.
a Find, at the 2.5% level, the critical region for this test, stating your hypotheses clearly.
The mean score on the test for the sample of 80 people was 102.5.
b Comment on this observation in light of the critical region.
X~N(100, 152)
inverse normal
area: 0.025
N = 0 0 = 1
b) 102.54103.29
-1.959
d = -2.5758 (0.015) + 0.58 = 0.574
O
"
-2.5758
1.959
-2.5%
→not sufficent evidence
to reject Ho
→ cannot say chocolate improves 10.
с
1.959= X-100
15
√80
writical region X > 103.29
0.005
2.5758
X = 1.959 (150) +100
=103.285

Normal Distributional

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Normal Distributional

Can't find what you're looking for? Explore other subjects.

Knowunity is the # 1 ranked education app in five European countries

Knowunity was a featured story by Apple and has consistently topped the app store charts within the education category in Germany, Italy, Poland, Switzerland and United Kingdom. Join Knowunity today and help millions of students around the world.

4.9+

13 M

#1

950 K+

Still not sure? Look at what your fellow peers are saying...

I love this app so much [...] I recommend Knowunity to everyone!!! I went from a C to an A with it :D

The application is very simple and well designed. So far I have found what I was looking for :D

Love this App ❤️, I use it basically all the time whenever I'm studying