Download What is the probability of a normal random variable taking a value

Ex: What is the probability of a normal random variable taking a value
between 1 s.d below and 1 s.d. Above its mean P(-1 < Z < 1)? P( -­‐1 < Z < 1)=P(Z < 1)-­‐P(Z < -­‐1)=0.8413-­‐0.1587=0.6826 -­‐1 1 Ex: The probability is 0.01 that a standardized normal random variable
takes a value below what particular value of z? The closest value to 0.01 is 0.0099, corresponding to z=-2.33
Ex: The probability is 0.15 that a standardized normal random variable
takes a value above what particular value of z? 0.15 ? Method 1: by symmetry, P(Z > ?)=P(Z < -?)=0.15
Normal Table
01/03/12 18:53
-1.1
.1357
.1335
.1314
.1292
.1271
.1251
.1230
.1210
.1190
.1170
-1.0
.1587
.1562
.1539
.1515
.1492
.1469
.1446
.1423
.1401
.1379
-0.9
.1841
.1814
.1788
.1762
.1736
.1711
.1685
.1660
.1635
.1611
-0.8
.2119
.2090
.2061
.2033
.2005
.1977
.1949
.1922
.1894
.1867
?=-­‐1.04 -0.7
.2420
.2389
.2358
.2327
.2296
.2266
.2236
.2206
.2177
.2148
-0.6
.2743
.2709
.2676
.2643
.2611
.2578
.2546
.2514
.2483
.2451
-0.5
.3085
.3050
.3015
.2981
.2946
.2912
.2877
.2843
.2810
.2776
Method 2 P(Z > ?) => 1-P(Z < ?)= 1-0.15=0.85 probability of being below the
value we seek.
? = 1.04
What is the probability of a normal variable being lower than 5.2 standard
deviations below its mean?
P(Z < -5.2) = 0
(This value (-5.2) is not in the table, but we know that P(Z < -3)=0.0015)
What is the probability of a normal random variable being higher than 6
standard deviations below its mean?
P(Z > -6) = 1-P(Z < -6) = 1-0 = 1
MALE FOOT LENGTH “RIVISITED”
Male foot length have a normal distribution with µ=11 and σ=1.5.
a) What is the probability of a foot length of more than 13 inches?
First we standardize the value of 13 inches:
x − µ 13 −11
=
= 1.33
σ
1.5
P(X > 13)=P(Z > 1.33) = by symmetry = P(Z < -1.33) =0.0918
z=
€
b) What is the probability of a male foot length between 10 and 12
inches?
We have to standardize the values of 10 and 12 inches:
z1 =
10 −11
= −0.67
1.5
z2 =
12 −11
= 0.67
1.5
P(10 < X < 12) = P(-0.67 < Z < 0.67)
=P(Z < 0.67)-P(Z < -0.67)=0.7486-0.2514=0.4972
€
Ex.: According to the College Board website, the scores of a math exam in a
certain year had a mean of 507 and a s.d of 111. Assume that the scores
follow a normal distribution.
a) What is the probability that a randomly chosen student scored above 700?
z=
700 − 507
= 1.738 ≅1.74
111
P(X > 700) = P(Z > 1.74) = by symmetry = P(Z < -1.74) = 0.0409
€
b) What proportion of students score between 400 and 600 in the math
exam?
400 − 507
z1 =
= −0.96
111
600 − 507
z2 =
= 0.837 ≅ 0.84
111
P( 400 < X < 600) = P( -0.96 < Z < 0.84) =
€
= P(Z < 0.83) – P(Z < -0.96) = 0.7995-0.1685 = 0.631
FROM P to X
The probability is 0.04 that a randomly chosen adult male foot length will
be less than how many inches ?
The closest value on the table is 0.0401, associated to z=-1.75
z = −1.75 =
€
x −11
⇒ x = 11 − (1.75)(1.5) = 8.375
1.5
Ex.: The probability is 0.1 that a randomly chosen adult male foot length
will be longer than how many inches ?
P(Z > 1.28) = P(Z < -1.28)=0.10
1.28 =
€
x −11
⇒ x = 11+ (1.28)(1.5) = 12.92
1.5
Ex.: A study reported that the amount of money spent each week for lunch
by a worker in a particular city is a normal random variable with mean $35
and s.d $5.
a) The probability is 0.97 that a worker will spend less than how
money in a week for lunch?
much
b) There is a 30% chance of spending more than how much for lunch in a
week?
a)
P(Z < 1.88)=0.9699
1.88 =
€
x − 35
⇒ x = 35 + (1.88)(5) = 44.4
5
b) There is a 30% chance of spending more than how much
for lunch in a week?
P(Z > 0.52) = P(Z < -0.52) = 0.3015
x − 35
0.52 =
⇒ x = 35 + (0.52)(5) = 37.6
5
€
A certain type of storage battery lasts, on average, 3.0 years with a
standard deviation of 0.5 year. Assuming that the battery lives are
normally distributed, find the probability that a given battery will last less
than 2.3 years.
2.3 − 3.0
= −1.40
0.5
P(X < 2.3) = P(Z < −1.40) = 0.0808
z=
An electrical
firm manufactures light bulbs that have a life, before burn€
out, that is normally distributed with mean 800 hours and a standard
deviation of 40 hours. Find the probability that a bulb burns between 778
and 834 hours.
778 − 800
834 − 800
= −0.55 z2 =
= 0.85
40
40
P(778 < X < 834) = P(−0.55 < Z < 0.85) =
P(Z < 0.85) − P(Z < −0.55) =
= 0.8023 − 0.2912 = 0.5111
z1 =
Gauges are used to reject all components where a certain dimension is not
within the specification 1.50±d.
It is known that this measurement is normally distributed with mean 1.50
and s.d. 0.2. Determine d such that the specifications “cover” 95% of the
measurement.
95%  Area uder µ-2σ and µ+2σ
P(-1.96 < Z < 1.96)=0.95
1.96 =
€
(1.50 + d) −1.50
⇒ d = (0.2)(1.96) = 0.392
0.2
NORMAL APPROXIMATION TO BINOMIAL
Under certain conditions the normal distribution can provide a very good
approximation to the binomial distribution.
Ex: Suppose a student answers 20 true/false questions completely at
random. What is the probability of getting no more than 8 correct? Let
X be the number of questions the student gets right (successes) out of
the 20 questions (trials), when the probability of success is 0.5. X is
therefore binomial with n=20 and p=0.5.
8
P(X ≤ 8) = ∑ b(x;20,0.5)
x =0
€
If we can not calculate it by hand, and
we have not tables or any software,
we can use the normal distribution.
The binomial is a symmetric distribution, it can be approximated by a
normal with:
µ = np
σ = np(1 − p)
µ = np = (20)(0.5) = 10 σ = 2.24
⎛
8 −10 ⎞
P(X B ≤ 8) ≈ P(X N ≤ 8) = P⎜ Z <
⎟ = P(Z < −0.89) = 0.1867
⎝
2.24 ⎠
BUT IF WE USE THE TRUE BINOMIAL P(X ≤ 8)=0.2517
€
PROBLEM: the approximation is good for larger n
“Rule of thumb”: for having a good approximation np ≥ 10 and n(1-p) ≥ 10
n=20
n=200
n=40
n=2000
IMPROVING THE APPROXIMATION: CONTINUITY CORRECTION
Suppose we want to
calculate P(X ≤ 20).
It corresponds to
the area under the
curve up to 20. But
in the histogram
(discrete values) we
have to consider
the area up to 20.5
to consider all the
histogram bar.
We can improve the approximation calculating the probability
that the normal variable is X ≤ 20.5.
Applying the continuity correction in the previous exercise:
µ = np = (20)(0.5) = 10 σ = 2.24
⎛
8.5 −10 ⎞
P(X B ≤ 8) ≈ P(X N ≤ 8.5) = P⎜ Z <
⎟ = P(Z < −0.67) = 0.2514
⎝
2.24 ⎠
TRUE BINOMIAL P(X ≤ 8)=0.2517
€
Ex.: Roughly 10% of all college students in US are left-handed. Most
academic institutions, therefore, try to have at least few lefthanded chairs in each classroom. 225 students are about to enter a
lecture hall that has 30 left-handed chairs for a lecture. What is the
probability that this is not going to be enough, i.e. That more than 30
students are left-handed?
Binomial with n=225 p=0.1 P(X > 30)=P(X ≥ 31)=?
“Rule of thumb”: np=22.5 n(1-p)=202.5
OK
Using the continuity correction:
⎛
30.5 − 22.5 ⎞
P(X B ≥ 31) = P(X N ≥ 30.5) = P⎜ Z ≥
⎟ = P(Z ≥1.78) = P(Z ≤ −1.78) = 0.0375
⎝
⎠
4.5
€