Week 3 – SS3 First Term Further Mathematics Notes

WEEK THREE
PROBABILITY DISTRIBUTION (CONTINUATION)

• Normal distribution
• properties and area
• z – scores application.

  Normal Distribution: This is a continuous distribution and it takes the form
  P(x) = 1 e^{-½ (x – u)2}
  ð√2x^ð
  
   where∂ is the standard deviation. U is the mean and e = 2.718.
  The graphical representation of a normal distribution is a bell-shaped curve.

   

P(x)

 
 
 
 
 
 
 u
PROPERTIES OF THE NORMAL DISTRIBUTION

1. It depends on the mean (u) and standard deviation
2. The shape is bell-shaped
3. The function is continuous, hence the range is from –ά to + ά
4. The curve is symmetrical about the vertical line through the mean.

 A normal distribution function is a probability function, hence the total area under the curve is 1 .
The normal distribution has a complicated equation, but it can be shown in shaded area under the shape

1. Values within 1 sd of the mean

 
 
 Pr (μ – r <x ≤µ + r) = 0.68.

 
 
 x
µ-ðµ µ+ð
2. Values within 2 sd of the mean

    

 
 
 
     Pr(μ-2r <x < μ+ 2R) = 0.955

 
      µ– 2r µ µ+2r
3. Values within 3sd of the mean.

 
 
 
 Pr(μ-3r <x < μ + 3R) = 0.997

 
  u– 2r µ µ+2r

 Example: A random variable X is normally distributed with mean 65 and standard deviation 5, find:

1. Pr (60 ≤ x≤ 70) II. Pr (55 ≤ x ≤ 75) III. Pr(50 ≤ x ≤80 )

Solution:
µ = 65, r = 5
1. Pr (60 ≤ x ≤ 70)     =    Pr (µ – r ≤ x ≤ µ + r)
                =    Pr (65-5≤x ≤ 65+5)
                =    Pr (60 ≤ x ≤ 70) = 0.68
II. Pr (55 ≤ x ≤ 75)     =    Pr (µ -2r ≤ x ≤ µ + 2r)
                =    0.95
III. Pr (50 ≤ x ≤ 80)     =    Pr (µ -35 ≤ x ≤ µ +3 r)
                = 0.997

 EVALUATION
1. A random variable x is normally distributed with mean 45 and standard deviation 12.
Find:     I Pr (9 < x <81) II. Pr(33 < x < 57)

 AREA UNDER NORMAL CURVE
The area under a normal curve can be defined by checking the probability value in the Normal distribution probabilities table.
Example:

1. Find the area between Z = 0 and Z = 2.13

Solution:
Pr( 0 < z < 2.13) check the value against 2.13.
Pr( 0 < z < 2.13) = 0.4834

  0 2.13
2. Find the area z = -1.3 and z = 1.2

 Solution
Pr (-1.3 < z<1.2)
= Pr (0 < z<1.2) +Pr (0 < z<1.3)
= 0. 3849 + 0.4032.
= 0.7881.

 3. Find the area between Z = 0.36 and Z = 1.89

 Solution
Pr (0.36 < z<1.89)
=Pr (0 < z<1.2) – Pr (0 < z<0.36)
= 0.4706 – 0.1406
= 0.33.

 EVALUATION
Using the standard deviation normal distribution table. Find the area under
1. Pr (-1.5 < z< 2.0) 2.Pr(z <1.5) 3.Pr (z >2.6)

 Z Scores
The area under a normal distribution curve between two values depends on the number of standard deviations from the mean. Therefore, the standardize normal curve is obtained from the normal curve by the substitution.
Z = X – µ
σ
:. Z is called the standardized score or Z score at mean zero (o) and standard deviation 1.

 Example: A random variable whose distribution is normal has mean 25 and standard deviation 5. Find
I. Pr (22 < z< 27) II.Pr (x <20) III.Pr (x > 26.5)
Solution
    µ= 25, = 5
I. Pr(22 < x < 27) = Pr ( x₁-µ<z <x₂ – µ)
σσ =Pr(z₁< z< z₂)
Z₁= 22 – 25 = -3= – 0.6
5 5
Z₂ = 27-25 = ²/₅ = 0.4
5
Pr (-1.3 < z<1.2) = Pr (-0.6 < z<0.4)
= Pr (z <0.4) + Pr (z<0.6)
= 0.1554 + 0.2258 = 0.3812
II. Pr(x <20) = Pr (z< 20-25)
5
=Pr (z<-1) = Pr (-1 < z<0)
=Pr (z< 0) – Pr (z<1)
= 0.5 – 0.3413.
:. Pr(x <20) = 0.1587
III. Pr(x > 26.5) = = Pr (z>26.5 – 25)
5
         = Pr (z > 0.3) = Pr (0.3 < z < ∞)
         =Pr (z < ∞) – Pr (z < 0.3)
     = 0.5 – 0.1179
= 0.3821.

 EVALUATION
The weights of packets of sugar produced by a machine have a mean of 1kg and a standard deviation of 0.1kg. What is the probability that in a random sample of 50 packets the combined weight will exceed 52kg?

 GENERAL EVALUATION

1. The inner diameters of bolts produced in a factory are normally distributed with mean 5cm and standard deviation 0.02cm. Find (a) the percentage of the number of bolts with inner diameters less than 5.015cm; (b) the probability that a bolt will have an inner diameter between 4.995cm and 5.015cm.
2. Use the standard normal distribution table to find (i) Pr (Z > 2.6) (ii) Pr (- 1.5 < Z < 1.7)

 READING ASIGNMENT
Read Z scores and Normal distribution. Further Mathematics Project III Page 2 202-210.

 WEEKEND ASSIGNMENT
1.    Find the area between z = 0.36 and z= 1.89 (a) 0.33    (b) 0.6112    (c) 1.00
Use the information below to answer questions 2-4.
A distribution with mean 85 and standard deviation 10 is normally distributed. If x is a random variable of the distribution, find
2.    Pr (80 < x <8.9) (a) 0.9332    (b) 0.5     (c) 0.3469
3.    Pr(x >83)        (a) 0.1587    (b) 0.789    (c) 0.4207
4.    Pr(x > 87)         (a) 0.0047    (b) 0.35    (c) 0.4207
5.    Find, with the usual notations, P (z <1.810) from the table of normal distribution.
    (a) 0.311        (b) 0.0288    (c) 0.9649

 THEORY

1. The scores of some 500 candidates in an examination were found to be approximately normally distributed with mean 40 and standard deviation 5. Find the number of candidates who scored at least 48.
2. The lengths of nails produced in a factory are approximately normally distributed with mean 2cm and a standard deviation 0.01cm. Find the proportion of nails that will be shorter than 1.98cm.