8.2b Standard Normal Distribution Tables (Example 2)


8.2b Standard Normal Distribution Tables (Example 2)

Example 2:
Find the value of each of the following probabilities by reading the standard normal distribution tables.
(a) P (0.4 < Z < 1.2)
(b) P (–1 < Z < 2.5)
(c) P (–1.3 < Z < –0.5)

Solution:
(a)
P (0.4 < Z < 1.2)
= Area P – Area Q
= Q (0.4) – Q (1.2) ← (reading from the standard normal distribution table for 0.4 and 1.2 are 0.3446 and 0.1151 respectively)
= 0.3446 – 0.1151
= 0.2295









(b)
P (–1 < Z < 2.5)
= 1– Area P – Area Q
= 1 – Q (1) – Q (2.5)
= 1 – 0.1587 – 0.00621 ← (reading from the standard normal distribution table for 1 and 2.5 are 0.1587 and 0.00621 respectively)
= 0.8351


(c)
P (–1.3 < Z < –0.5)
= Area P – Area Q
= Q (0.5) – Q (1.3)
= 0.3085 – 0.0968 ← (reading from the standard normal distribution table for 0.5 and 1.3 are 0.3085 and 0.0968 respectively)
= 0.2117









8.2c Probability of an Event


8.2c Probability of an Event

Example:
The masses of pears in a fruit stall are normally distributed with a mean of 220 g and a variance of 100 g. Find the probability that a pear that is picked at random has a mass
(a) of more than 230 g.
(b) between 210 g and 225 g.
Hence, find the value of h such that 90% of the pears weigh more than h g.

Solution:
µ = 220 g
σ = √100 = 10 g
Let X be the mass of a pear.

(a)
P(X>230)=P(Z>23022010)Convert to standard normal distribution using Z=Xμσ=P(Z>1)=0.1587


(b)
P(210<X<225)=P(21022010<Z<22522010)Convert to standard normal distribution using Z=Xμσ=P(1<Z<0.5)=1P(Z>1)P(Z>0.5)=10.15870.3085=0.5328

For 90% (probability = 0.9) of the pears weigh more than h g, 
(X > h) = 0.9
(X < h) = 1 – 0.9
= 0.1

From the standard normal distribution table,
(Z > 0.4602) = 0.1
(Z < –0.4602) = 0.1
h22010=0.4602h220=4.602h=215.4



8.2b Standard Normal Distribution Tables (Example 3)


8.2b Standard Normal Distribution Tables (Example 3)

Example 3:
Find the value of k if
(a) (Z > k) = 0.0480
(b) (Z > k) = 0.8350

Solution:
(a)


From the standard normal distribution table, k = 1.665

Z
6
5 (Subtract)
1.6
.0485
5

(b)

 
From the standard normal distribution table,
k –0.974 ← (Remember to put a negative sign at the value of k because it is on the left-hand side of the normal curve.)

Z
7
4 (Subtract)
0.9
.1660
10

8.2b Standard Normal Distribution Tables (Example 1)


8.2b Standard Normal Distribution Tables (Example 1)

Example 1:
Find the value of each of the following probabilities by reading the standard normal distribution tables.
(a) (Z > 0.600)
(b) (Z < –1.24)
(c) (Z > –1.1)
(d) (Z < 0.76)
 
Solution:
Standard Normal Distribution Table


*When reading the standard normal distribution tables, it involves subtraction of values.

(a)
From the standard normal distribution table,(Z > 0.600) = 0.2743


(b)
(Z < –1.24)
= (Z > 1.24)
= (1.24)
= 0.1075 ← (reading from the standard normal distribution table)


(*In the standard normal distribution table, all the values of are positive. As the curve is symmetrical about the vertical axis, the area of the shaded region in both of the graphs are the same.)


(c)
(Z > –1.1)
= 1 – Area P
= 1 – (–1.1)
= 1 – 0.1357 ← (reading from the standard normal distribution table)
= 0.8643

(d)
(Z < 0.76)
= 1 – Area P
= 1 – (0.76)
= 1 – 0.2236 ← (reading from the standard normal distribution table)
= 0.7764


4.3.1a Addition of Vectors (Examples)

Example 1:

Find
(a) The resultant vector of the addition of the two parallel vectors above
(b) The magnitude of the resultant vector.

Solution:
(a)
Resultant vector
= addition of the two vectors
=PQ+RS

(b)
Magnitude of the resultant vector
=|PQ|+|RS|=|62+82|+|62+82|=10+10=20units



Example 2:

The diagram above shows a parallelogram OABC. M is the midpoint of BC. Vector OA=a˜ and OC=c˜.  Find the following vectors in terms of a˜ and c˜.
(a)OB(b)MB(c)OM
 
Solution:
(a)
OB=OA+ABTriangle Law of addition=OA+OCAB=OC=a˜+c˜

(b)

MB=12CBM is the midpoint of CB =12OA =12a˜

(c)
OM=OC+CMTriangle Law of addition=OC+MBCM=MB=c˜+12a˜

3.6.1 Integration as the Summation of Volumes – Examples

Example 1:
Find the volume generated for the following diagram when the shaded region is revolved through 360° about the x-axis.


Solution:
Volume generated, Vx
Vx=πbay2dxVx=π42(3x8x)2dxVx=π42(3x8x)(3x8x)dxVx=π42(9x248+64x2)dxVx=π[9x3348x+64x11]42Vx=π[3x348x64x]42Vx=π[(3(4)348(4)644)(3(2)348(2)642)]Vx=π(16+104)Vx=88πunit3


Example 2:
Find the volume generated for the following diagram when the shaded region is revolved through 360° about the y-axis.


Solution:
Volume generated, Vy
Vy=πbax2dyVy=π21(2y)2dyVy=π21(4y2)dyVy=π214y2dyVy=π[4y11]21=π[4y]21Vy=π[(42)(41)]Vy=2πunit3


3.5.1 Integration as the Summation of Areas – Examples


Example 1
Find the area of the shaded region.


Solution:
Area of the shaded region=baydx=40(6xx2)dx=[6x22x33]40=[3(4)2(4)33]0=2623unit2


Example 2
Find the area of the shaded region.


Solution:
y = x -----(1)
x = 8yy2-----(2)
Substitute (1) into (2),
y = 8yy2
y2 – 7y = 0
y (y – 7) = 0
y = 0 or 7
From (1), x = 0 or 7
Therefore the intersection points of the curve and the straight line is (0, 0) and (7, 7).

Intersection point of the curve and y-axis is,
x = 8yy2
At y-axis, x = 0
0 = 8yy2
y (y – 8) = 0
y = 0, 8

Area of shaded region = (A1) Area of triangle + (A2) Area under the curve from y = 7 to y = 8.
=12×base×height +87xdy=12×(7)(7)+87(8yy2)dy=492+[8y22y33]87=2412+[4(8)2(8)33][4(7)2(7)33]=2412+85138123=2816unit2

3.4b Laws of Definite Integrals


3.4b Laws of Definite Integrals



Example:
Given that 73f(x)dx=5 , find the values for each of the following:

(a)736f(x)dx(b)73[3f(x)]dx(c)372f(x)dx(d)43f(x)dx+54f(x)dx+73f(x)dx(e)73f(x)+72dx


Solution:
(a)736f(x)dx=673f(x)dx=6(5)=30(b)73[3f(x)]dx=733dx73f(x)dx=[3x]735=[3(7)3(3)]5=7(c)372f(x)dx=732f(x)dx=273f(x)dx=2(5)=10(d)43f(x)dx+54f(x)dx+73f(x)dx=73f(x)dx=5(e)73f(x)+72dx=73[12f(x)+72]dx=7312f(x)dx+7372dx=1273f(x)dx+[7x2]73=12(5)+[7(7)27(3)2]=52+14=1612


Short Questions (Question 10)


Question 10 (4 marks):
Diagram 10 shows the position of three campsites A, B and C at a part of a riverbank drawn on a Cartesian plane, such that A and B lie on the same straight riverbank.

Diagram 10

Sam wants to cross the river from campsite C to the opposite riverbank where the campsites A and B are located.
Find the shortest distance, in m, that he can take to cross the river. Give your answer correct to four decimal places.

Solution:

Let the shortest distance from campsite Cto opposite riverbank is CD.Area of ABC=12|2     7     4     3     5   3  2  3|=12|(2)(5)+(7)(3)+(4)(3)(3)(7)(5)(4)(3)(2)|=12|66|=33 units2Distance of AB=(27)2+(35)2=85 mArea of ABC=3312(AB)(CD)=3312(85)(CD)=33CD=33(2)85CD=7.1587 (4 d.p.)Thus, the shortest distance is 7.1587 m.