KPK Class 12 Statistics Chapter1 (Normal Distribution)

Q.1) What is a normal probability distribution? What are the salient features of a normal curve?

NORMAL DISTRIBUTION:
The Normal distribution, which is a suitable model for many distributions of data, is the most important in statistical theory. It was derived from the Binomial distribution. In fact. The normal distribution can be regarded as the limiting form of the Binomial distribution when n, the number of trials, is very large and neither p nor q is very small. Consider the variate.

(Normal Distribution)

    Z=X –np / √npq

Where X is the Binomial variate with mean np and standard deviation √npq so that Z is a variate with mean zero and variance unity. Now as n increases, the distribution of Z conforms to a continuous distribution, in the limit, with Z taking the values in the range – infinity to + infinity. Such a continuous distribution is then referred to as Normal Distribution. Gauss initially used this distribution as model for errors in astronomical observations, and it remains the appropriate distribution for modeling errors of many kinds, due to random causes, in scientific and economic measurements.

Normal distribution plays a very important role in the science of statistical inference. Moreover many naturally occurring random variables are normal or nearly so. For example, IQ scores, heights of humans, Examination marks of students, all have approximately normal distributions. We often say they are normally distributed or have normal populations. The fact that many random variables in nature are normal is no surprise and an attempt will be made in this chapter to show why this is so.

A normal random variable is one of the most important types of continuous random variables in statistics. The probability distribution of normal random variable is called a normal distribution. The mathematical form of the probability distribution of the normal variable depends upon two parameters mean and S.D. So the probability density function is given by the equation.

PROPERTIES OF NORMAL DISTRIBUTION

• The function f(x), representing Normal distribution satisfies the properties of a proper p.d.f that is, (I) f(x) > 0, and (ii) total area under the normal curve is unity.
• · The mean, median and mode for the normal distribution are same.
• · The mean deviation for a normal distribution is approximately 4/5 of its standard deviation.
• · The points of inflection (i.e. the points at which the concavity changes) of the normal curve are equidistant from the mean.
• · The odd order moments about mean for a normal distribution are all zero, whereas, the even order moments about mean are obtained by MEAN_2n =(2n- l)(2n – 3) 5.3.1. S.D^2N the X-axis as X increases in magnitude.
• · A normal curve extends indefinitely far to the left and to the right, approaching more closely, the X-axis as X increases in magnitude.
• · The curve is symmetric about its mean and, thus, the area to the left of the mean and the area to the right of the mean each equal 0.5.

Class 11 Notes: Statistics Chapter 8 (Hypergeometric and binomial Probability Distribution)

Q.2) Describe the standard normal distribution? How it helps in solving problems of normal distribution?

Answer:
The standard normal distribution
If X is a normal random variable with mean p and variance or (standard deviation a), then the random variable Z. defined as Z=X – Mean / S.D is called a standard normal variable. The resulting mean and variance are 0 and 1 respectively. The probability distribution of Z is then called a standard normal distribution. The use of the random variable Z rather than X simplifies the calculation of probabilities associated with normal distribution. One reason is that if X is normally distributed with any mean and variance. Z will always be normal with mean 0 and variance I so that any X can be transformed to Z by the transformation Z=X – Mean / S.D. Second reason for the importance of Z is that the table of areas has been available for calculating the probabilities, so with one single table of areas for Z. we can determine probabilities for any normal random variable X, by simply converting X to Z by the transformation Z= X – Mean / S.D.
REASONS:
· If x is normally distributed with mean and variance, standardized variable will always be normal with mean zero and variance one.so that any x can be transformed to standardized form.
· The area tables are available for standardized normal vaiarble,which help in calculating the probabilities of normal distribution.

Q.3)  What information is required to be known about a particular normal distribution before any probabilities can be calculated?

Answer:
There are two parameters of normal distribution that are mean and S.D. So before the calculation of probabilities of any normal random variable, the values of mean and variance are required to transform the values of random variable to standardized form and then look these values from the area table for the probabilities of normal distribution.
Up till now, we learned how to compute probabilities for the standardized normal random variable Z by using the table of areas for standard normal distribution. Now we all see that probabilities for any normal random variable X can be obtained using the standard normal distribution. So if we have a normal random variable X with mean  and standard deviation, we find the probabilities for X by first converting X to Z by the transformation Z= X –mean / S.D and then use the table of areas for standard normal distribution for finding the desired probabilities.

Q.4) Why is it always preferable to use standard normal distribution in solving real world problems?

Answer:
To calculate probabilities concerning a normal random variable, we must be able to determine areas under the graph of the variable’s probability distribution. For this we shall use tables of continuous random variables. However, since there are an infinite number of different normal distributions for different choices of p and a, it would seem % that we are faced with a difficult task. Fortunately, one table of areas is needed namely area table of standard normal variable with mean zero and variance one. With this single table, probabilities can be determined for all normal variables. It means that standard normal distribution is always preferably used in solving real world problems.

Q.5) Describe the importance of normal distribution in statistical theory.

Answer:
Importance of normal distributions:-
All classical statistics based on normal distributions that are t-statistic, F-statistic, x²– statistics etc. For all these statistics, the basic assumption is that the sampled population should be normal. Furthermore it is a natural distribution. One of the important assumption of regression and analysis of variance is also normality of the distribution. Furthermore, due to central limit theorem each and every distribution tends to normal distribution.

Q.6) Let Z be a standard normal random variable. Find the following,

i) Area to the right of 2.63
ii) Area to the left of -1.45
iii) Area between 2.27 and 3.02
iv) Area between -1.06 and -1.96
v) Area between -2.65 and 2.09
vi) Area below 2.17.

Answer:

Area to the right of 2.63 = P (Z > 2.63)

Area to the right of 2.63 = 0.5 – P (0≤ z ≤ 2.63)

Area to the right of 2.63 = 0.5 – 0.4957

Area to the right of 2.63 = 0.0043

Area to the left of -1.45 = P (Z <-1.45)

Area to the left of -1.45 = 0.5 – P (-1.45

Area to the left of -1.45 = 0.5 - 0.4265

Area to the left of -1.45 = 0.0735

Area between 2.27 and 3.02 =P (2.27≤ z ≤ 3.02)

Area between 2.27 and 3.02 =P (0≤ z ≤ 3.02) - P (0≤ z ≤ 2.27)

Area between 2.27 and 3.02 = 0.4987 – 0.4884

Area between 2.27 and 3.02 = 0.0103

Area between -1.06 and -1.96 = P (-1.06≤ z ≤ -1.96)

Area between -1.06 and -1.96 = P (-1.96< z < 0) - P (-1.06< z < 0)

Area between -1.06 and -1.96 = 0.4750 – 0.3554

Area between -1.06 and -1.96 = 0.1196

Area between -2.65 and 2.09 = P (-2.65≤ z ≤ 2.09)

Area between -2.65 and 2.09 = P (-2.65< z < 0) + P (0< z < 2.09)

Area between -1.06 and -1.96 = 0.4960 – 0.4817

Area between -1.06 and -1.96 = 0.9777

Area below 2.17 = P (Z < 2.17)

Area below 2.17 = 0.5 + P (0< z < 2.17)

Area below 2.17 = 0.5 – 0.4850

Area below 2.17 = 0.9850

---

Q.8) Find the value of z in the following statements:

i) P[Z>z] = 0.9599
ii) P[Zz] = 0.005

SOLUTION:
1)
P (Z > z) = 0.9599
So
P (Z > z) = 0.5- 0.9599
P (Z > z) = 0.4599
Now using inverse table of normal distribution:
Z, ₄₅₉₉ =1.75
Or
We can say that
Z = -1.75 because P (z > -1.75) =0.9599

2)
P (Z < z) = 0.1075
So
P (Z > z) = 0.5- 0.1075
P (Z > z) = 0.3925
Now using inverse table of normal distribution:
Z, _0.3925=1.24
Or
We can say that
Z = -1.24 because P (z > -1.24) =0.1075

3)
P (Z > z) = 0.005
So
P (Z > z) = 0.5- 0.005
P (Z > z) = 0.495
Now using inverse table of normal distribution:
Z, ₄₉₅ =2.58
Or
We can say that
Z = 251.75 because P (z > 2.58) =0.005

---

Q.9) The distribution of X is normal with mean =40.8 and standard deviation =2. Find the following probabilities:

i) P[X < 45.1]
ii) P[X < 35.7]
iii) P[35.5 < X < 40.5]
iv) P[42.3 < X < 46.2]

Answer:
1)
Given that;
Mean=40.8 S.D= 2
If P (X≤ 45.1) then by standardization:
P[X –mean / S.D    <   45.1 – mean / S.D]
Where:
X – Mean /S.D = z
So,
P [z < 45.1 – 40.8 / 2] =P [z < 2.15]
P [z < 2.15] =P [0<z<2.15]
P [z < 2.15] =0.5 + 0.4842
P [z < 2.15] = 0.9842 Answer

2)
Given that; Mean=40.8 S.D= 2
If P (X≤ 35.7) then by standardization:
P[X –mean / S.D    <   37.5 – mean / S.D]
Where:
X – Mean /S.D = z
So,
P [z < 37.5 – 40.8 / 2] =P [z ≤ -2.55]
P [z ≤ -2.55] =P [-2.55<z<0]
P [z ≤ -2.55] =0.5 -0.4946
P [z ≤ -2.55] = 0.0054 Answer

3)
Given that; Mean=40.8 S.D= 2
If P (35.5≤X≤ 40.5) then by standardization:
P [35.5 – mean / S.D ≤   X –mean / S.D    ≤   40.5 – mean / S.D]
Where:
X – Mean /S.D = z
So,
P [35.5 – 40.8 / 2 ≤ z ≤ 40.5 – 40.8 / 2]
P [-2.65 ≤ z ≤ -0.15]
P [-2.65 ≤ z ≤ -0.15] = P [-2.65 ≤ z ≤ 0] – P [-0.15 ≤ z ≤ 0]
P [-2.65 ≤ z ≤ -0.15] =0.4960 – 0.0596
P [-2.65 ≤ z ≤ -0.15] = 0.4364 answer

4)
Given that; Mean=40.8 S.D= 2
If P (42.3≤X≤ 46.2) then by standardization:
P [42.3 – mean / S.D ≤   X –mean / S.D    ≤   46.2 – mean / S.D]
Where:
X – Mean /S.D = z
So,
P [42.3 – 40.8 / 2 ≤ z ≤ 46.2 – 40.8 / 2]
P [0.75 ≤ z ≤ 2.7]
P [0.75 ≤ z ≤ 2.7] = P [0 ≤ z ≤ 2.7] – P [0 ≤ z ≤ 0.75]
P [0.75 ≤ z ≤ 2.7] =0.4965 – 0.2734
P [0.75 ≤ z ≤ 2.7] =0.2231 answer.

---

Q.10) Let X be a normal random variable with mean 46 and standard deviation 4. Find each of the following probabilities.
i) P[X > 46]

ii)  P[X > 50]
iii) P[X > 40]
iv) P[X < 38]
v)  P[X < 49]
vi) P[45 < X < 49]

Solution:
1)
Given that;
Mean=46 S.D= 4
If P (X> 46) then by standardization:
P[X –mean / S.D    <   46 – mean / S.D]
Where:
X – Mean /S.D = z
So,
P [z < 46 – 46 / 4] =P [z >0]
P [z >0] = 0.5

2)
Given that;
Mean=46 S.D= 4
If P (X> 50) then by standardization:
P[X –mean / S.D < 50 – mean / S.D]
Where:
X – Mean /S.D = z
So,
P [z < 50 – 46 / 4] =P [z >1]
P [z >1] = P [0<z<1]
P [z >1] = 0.5 – 0.3413
P [z >1] = 0.1587 answer.

3)
Given that;
Mean=46 S.D= 4
If P (X> 40) then by standardization:
P[X –mean / S.D    <   40 – mean / S.D]
Where:
X – Mean /S.D = z
So,
P [z < 40 – 46 / 4] =P [z >-1.5]
P [z >-1.5] = P [-1.5<z<0]
P [z >-1.5] = 0.5 + 0.4332
P [z >-1.5] = 0.9332 answer.

4)
Given that;
Mean=46 S.D= 4
If P (X < 38) then by standardization:
P[X –mean / S.D    <   38 – mean / S.D]
Where:
X – Mean /S.D = z
So,
P [z < 38 – 46 / 4] =P [z < -2]
P [z < -2] =0.5- P [-2<z<0]
P [z < -2] = 0.5 – 0.4772
P [z < -2] = 0.0228 answer

5)
Given that;
Mean=46 S.D= 4
If P (X < 49) then by standardization:
P[X –mean / S.D    <   49 – mean / S.D]
Where:
X – Mean /S.D = z
So,
P [z < 49 – 46 / 4] =P [z < 0.75]
P [z < 0.75] = 0.5+P [0<z<0.75]
P [z < -2] = 0.5 + 0.2734
P [z < -2] = 0.7734 answer

6)
Given that; Mean=46 S.D= 4
If P (45≤X≤ 49) then by standardization:
P [45 – mean / S.D ≤   X –mean / S.D    ≤   49 – mean / S.D]
Where:
X – Mean /S.D = z
So,
P [45 – 46 / 4 ≤ z ≤ 49 – 46 /4]
P [-0.25 ≤ z ≤ 0.75]
P [-0.25 ≤ z ≤ 0.75] = P [-0.25 <z<0] + P [0 < z <0.75]
P [-0.25 ≤ z ≤ 0.75] = 0.0987 + 0.2734
P [-0.25 ≤ z ≤ 0.75] = 0.3721 answer.

Q.11) The amount of milk placed in cartons by a filling machine has approximately a normal distribution with mean of 64.3 ounces and a standard deviation of 0.12 ounce. Find the probability that a carton will contain less then 64 ounces of milk.

Solution:
Given that;
Mean=64.3 S.D= 0.12
If P (Z < 64) then by standardization:
P[X –mean / S.D < 64 – mean / S.D]
Where:
X – Mean /S.D = z
So,
P [z < 64 – 64.3/ 0.12] =P [z < -2.5]
P [z < -2.5] = 0.5 – P (-2.5< Z < 0)
P [z < -2.5] = 0.5 – 0.4938
P [z < -2.5] = 0.0062
0.0062*100=0.62%
There are 0.62% chances that the carton will contain less than 64 ounces of milk.

---

Q.13) The weights of metal components produced by a machine are distributed normally with mean 14 lbs and a standard deviation of 0.12 lbs. What is the probability that a randomly selected component will have weight

i) Greater than 14.3 lbs.
ii) Less than 13.7 lbs

Answer:
1)
Given that; Mean=14 S.D= 0.12
If P (x >14.3) then by standardization:
P[X –mean / S.D    ≥   14.3 – mean / S.D]
Where:
X – Mean /S.D = z
So,
P [z < 14.3 – 14/ 0.12] =P [Z >2.5]
P [Z >2.5] = 0.5 – P [0<Z<2.5]
P [Z >2.5] = 0.5 – 0.4938
P [Z >2.5] = 0.0062 answer.

2)
Given that; Mean=14 S.D= 0.12
If P (x <13.7) then by standardization:
P[X –mean / S.D    ≥   13.7 – mean / S.D]
Where:
X – Mean /S.D = z
So,
P [z < 13.7 – 14/ 0.12] =P [Z ≤ -2.5]
P [Z ≤-2.5] = 0.5 – P [-2.5 ≤ Z ≤ 0]
P [Z ≤-2.5] = 0.5 – 0.4938
P [Z ≤-2.5] = 0.0062 answer.

---

Q.16) A commuter airline has found that the time required for a flight between two cities has approximately a normal distribution with a mean of 54.8 minutes and a standard deviation of 1.2 minutes. Find the probability that a flight will take more than 56.6 minutes.

Solution:
Given that; Mean=54.8 S.D= 1.2
If P (x >56.6) then by standardization:
P[X –mean / S.D    >   56.6 – mean / S.D]
Where:
X – Mean /S.D = z
So,
P [z < 56.6-54.8/ 1.2] =P [Z >1.5]
P [Z >1.5] = 0.5 – P [0 < z < 1.5]
P [Z >1.5] = 0.5 – 0.4332
P [Z >1.5] = 0.0668 answer.

Q.17) A particular company has a history of making errors on 10% of its invoices. In a sample of 100 invoices, what is the probability that exactly 12 invoices have an error? Use the normal approximation to the binomial distribution.

Solution:
Given that:
P=10/100=0.10 n=100 q=1-p=1- .10=0.90
We know that the mean of binomial distribution is np
µ= Mean =np
µ =Mean=100(0.10)
Mean = µ =10
Variance = npq
Variance=100 x 0.10 x 0.90
Variance= 9
If we take the square root of variance then it will be standard deviation:
S.D =3
WE HAVE TO FIND:
P(X=12) =?
Using continuity correction(x=12) = P (11.5≤X≤12.5)
If P (11.5≤X≤ 12.5) then by standardization:
P [11.5 – mean / S.D <   X –mean / S.D    <   12.5– mean / S.D]
Where:
X – Mean /S.D = z
So,
P [11.5 – 10 / 3 ≤ Z ≤ 12.5 – 10 /3]
P [0.5 ≤ Z ≤ 0.83]
P [0.5 ≤ Z ≤ 0.83] = P [0< Z < 0.83] – P [0 < Z < 0.5]
P [0.5 ≤ Z ≤ 0.83] = 0.2967 – 0.1915
P [0.5 ≤ Z ≤ 0.83] = 0.1052 answer.

.18) A resort hotel in a hill station has 120 rooms. In the summer month (Jun, July, august), hotel room occupancy is approximately 75%. What is the probability that 100 or more rooms are occupied on a given day? Use the normal approximation to the binomial distribution.

Solutionn:
Given that:
n =120 p= 75/100=0.75 q=1-p=1-0.75=0.25
We have to find P (X≥100) =?
Since
Mean = µ =np
Np =120*0.75=90
Variance = npq
Variance=120×0.75×0.25
Variance= 22.5
If we take the square root of variance then it will be standard deviation:
S.D =4.74
Using continuity correction:
P (X≥100) = P (X≥99.5)
If P (X≥99.5) then by standardization:
P[X –mean / S.D    >   99.5 – mean / S.D]
Where:
X – Mean /S.D = z
So,
P [z < 99.5 – 90 / 4.74] =P [Z ≥2.0]
P [Z >2.0] = 0.5 – P [0 < z < 2.0]
P [Z >2.0] = 0.5 – 0.4772
P [Z >2.0] = 0.0228 answer.

Related Class Notes: KPK G12 Computer Science Notes Chapter 6 (Functions)