1. The Normal Distribution
Chapter(6)
The Normal Distribution
Note: This PowerPoint is only a summary and your main source should be the book.

2. Introduction 6-1 Normal Distribution.
6-2 Applications of the Normal Distribution.
6-3 The Central Limit Theorem
Note: This PowerPoint is only a summary and your main source should be the book.

3. The Normal Distribution
6-1
The Normal Distribution
Note: This PowerPoint is only a summary and your main source should be the book.

4. (c) Positively skewed
(b) Negatively skewed
Normal
Mean =Median=Mode
Note: This PowerPoint is only a summary and your main source should be the book.

5. A normal distribution is a continuous , symmetric ,
bell shaped distribution of a variable.
A normal distribution curve depend on two parameters . σ
Position parameter
shape parameter
Note: This PowerPoint is only a summary and your main source should be the book.

6. The mathematical equation for the normal distribution: where e
population
mean
standard
deviation 2
.718 3
.14 ≈ π σ
Note: This PowerPoint is only a summary and your main source should be the book.

7. (1) Different means but same standard deviations.
Normal curves
with μ1 = μ2 and
σ1<σ2 2
(2) Same means but different standard deviations .
Note: This PowerPoint is only a summary and your main source should be the book.

8. (3) Different means and different standard deviations . 3
Note: This PowerPoint is only a summary and your main source should be the book.

9. Properties of the Normal Distribution
The normal distribution curve is bell-shaped.
The mean, median, and mode are equal and located at
the center of the distribution.
The normal distribution curve is unimodal (single mode).
The curve is symmetrical about the mean.
The curve is continuous.
The curve never touches the x-axis.
The total area under the normal distribution curve is
equal to 1 or 100%.
Note: This PowerPoint is only a summary and your main source should be the book.

10. The area under the normal distribution curve that lies within
one standard deviation of the mean is approximately 0.68 (68%). The interval between ( , )
two standard deviations of the mean is approximately 0.95 (95%). The interval between ( , )
three standard deviations of the mean is approximately (99.7%). The interval between ( , )
Note: This PowerPoint is only a summary and your main source should be the book.

11. (Areas Under the Normal Curve)
Empirical Rule: Normal Distribution
(Areas Under the Normal Curve)
Note: This PowerPoint is only a summary and your main source should be the book.

12. The Standard Normal Distribution
Note: This PowerPoint is only a summary and your main source should be the book.

13. The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1.
The formula for the standard normal distribution is
Note: This PowerPoint is only a summary and your main source should be the book.

14. All Normal Distribution can be transformed into standard Distribution.
Note: This PowerPoint is only a summary and your main source should be the book.

15. Empirical Rule: Standard Normal Distribution
Normal Distribution Curve
Standard Normal Distribution Curve
Note: This PowerPoint is only a summary and your main source should be the book.

16. The area under the standard normal distribution curve that lies within
one standard deviation of the mean is approximately 0.68 (68%). The interval between (-1,1).
two standard deviations of the mean is approximately 0.95 (95%). The interval between (-2,2).
three standard deviations of the mean is approximately (99.7%). The interval between (-3,3).

17. A).Finding Areas Under the Standard Normal Distribution Curve:
1. To the left of any Z value
P(z<-a) = P(z<a) = Q(a)
Note: This PowerPoint is only a summary and your main source should be the book.

18. P(z>-a) = P(z>a) = 1 - Q(a)
2.To the right of any Z value
P(z>-a) = P(z>a) = 1 - Q(a)
Note: This PowerPoint is only a summary and your main source should be the book.

19. P(-b<z<-a) = P(a<z<b) = P(z>a) – P(z>b)
3.Between any two Z values
P(-b<z<-a) = P(a<z<b) = P(z>a) – P(z>b)
= Q(a) – Q(b)
Note: This PowerPoint is only a summary and your main source should be the book.

20. Find the area to the left of z =1.99
Example :
Find the area to the left of z =1.99
P(Z<1.99)=0.9767
Example 6-1:
Find the area to the left of z =
Note: This PowerPoint is only a summary and your main source should be the book.

21. 1.9 9
Column
row
طريقة البحث بالجدول:
نبحث عن العددين ماقبل ومابعد النقطة من العامود
بينما العدد الثاني مابعد النقطة نبحث عنه من الصف
مثال:
1.99
2.09
Note: This PowerPoint is only a summary and your main source should be the book.

22. Find the area to the right of z = -1.16
Example 6-2:
Find the area to the right of z = -1.16
دائما نقوم بتحويل اشارة «أكبر من» إلى «أصغر من» كالتالي:
P(Z>a) =1- P(Z<a)
P(Z>-1.16)=1-P(Z<-1.16) = =
Note: This PowerPoint is only a summary and your main source should be the book.

23. طريقة البحث بالجدول:
1.16-

24. Find the area between z=1.68 and z=-1.37
Example 6-3:
Find the area between z=1.68 and z=-1.37
P(-1.37<Z<1.68)=P(Z<1.68)-P(Z<-1.37) = =
Note: This PowerPoint is only a summary and your main source should be the book.

25. A Normal Distribution Curve as a Probability Distribution Curve:
The area under the standard normal distribution curve can also be thought of as a probability .
Note: This PowerPoint is only a summary and your main source should be the book.

26. Find probability for each a) P(0<z<2.32) b) P(z<1.65)
Example 6-4:
Find probability for each
a) P(0<z<2.32)
b) P(z<1.65)
c) P(z>1.91)
a) P(0<Z<2.32)=P(Z<2.32)-P(Z<0) =
=0.4898
Note: This PowerPoint is only a summary and your main source should be the book.

27. c) P(Z>1.91)=1-P(Z<1.91) =1-0.9719 =0.0281
b) P(Z<1.65)=0.9505
c) P(Z>1.91)=1-P(Z<1.91) =
=0.0281
Note: This PowerPoint is only a summary and your main source should be the book.

29. B).Find the Z value that corresponds to given area1 Z 2
Look up the area in table E to find the z value
we cannot find the area in the table Z 3 4 Z
we cannot find the area in the table
Look up the area in table E to find the z value
Note: This PowerPoint is only a summary and your main source should be the book.

30. For exercises 40 to 45: find the Z value that corresponds to the given area:
0.0188
z = -2.08 46 43
0.9671
z = 1.84 Z Z Z
0.0239 44 45 Z
0.8962 =
z = 1.98 =
z = -1.26
Note: This PowerPoint is only a summary and your main source should be the book.

31. 5 6 Z a a Z
a = ----
z = look in the table E
– a = ----
z = look in the table E 42 40 Z
0.4175
0.4066 Z =
z = 1.32 =
z = -1.39
Note: This PowerPoint is only a summary and your main source should be the book.

33. Example 6-5:
Find the z value such that the area under the standard normal distribution curve between 0 and the z value is
=0.7123
Z=0.56
Note: This PowerPoint is only a summary and your main source should be the book.

34. Note: This PowerPoint is only a summary and your main source should be the book.

35. Questions ???
1) Find the z value to the left of the mean so that 98.87% of the area under the distribution curve lies to the right of it.
=0.0113
Z=-2.28
2) Find two z values so that 48% of the middle area is bounded by them.
1) = 0.52
2) 0.52/ 2=0.2600
3) Z = ± 0.64
3) Find two z values, one positive and one negative, that are equidistant from the mean so that the area in the two tails total the following values (5%). 1) 0.05/ 2=0.0250
2) Z = ± 1.96
Note: This PowerPoint is only a summary and your main source should be the book.

37. Applications of the Normal
6-2
Applications of the Normal
Distribution
Note: This PowerPoint is only a summary and your main source should be the book.

38. First Step: find the z , using the formula.
All normally distributed variables can be transformed into the standard normally distribution by using the formula for the standard score: or
to solve the application problems in this section, we must use that three steps.
First Step: find the z , using the formula.
Second Step: draw the figure and Shading the area.
Third Step : find the areas under ,using table E.
Note: This PowerPoint is only a summary and your main source should be the book.

39. Example : Solution: P(Z<0.47)=0.6808 Example 6-6: See page 329
A survey by the National Retail Federation found that women spend on average $ for the Christmas holidays. Assume the standard deviation is $ Find the percentage of women who spend less than $160. Assume the variable is normally distributed.
Solution:
Step 2 :Draw the figure
Step 1 : Find the z value .
Step 3 :Find the area ,using table E.
P(Z<0.47)=0.6808
Example 6-6: See page 329
68.08% of the women spend less than 160$ at Christmas time.
Note: This PowerPoint is only a summary and your main source should be the book.

40. a) Between 27 and 31 pounds per month
Example 6-7:
Each month, an American household generates an average of 28 pounds of newspaper for garbage or recycling. Assume the standard deviation is 2 pounds. If a household is selected at random, Find the probability of its generating.
a) Between 27 and 31 pounds per month
b) More than 30.2 pounds per month
Assume the variable is approximately normally distributed.
Solution (a) :
Step 1 : Find the two z value .
Note: This PowerPoint is only a summary and your main source should be the book.

41. P(-0.5<Z<1.5)=P(Z<1.5)-P(Z<-0.5) =0.9332-0.3085 =0.6247
Step 2 :Draw the figure
Step 3 :Find the area ,using table E.
P(-0.5<Z<1.5)=P(Z<1.5)-P(Z<-0.5) =
=0.6247
The probability that household generates between 27 and 31 pounds of newspapers per month is 62.47%
Note: This PowerPoint is only a summary and your main source should be the book.

42. P(Z>1.1)= 1-P(Z<1.1) =1-0.8643 =0.1375 Solution (b) :
Step 1 : Find the z value .
Step 2 :Draw the figure
1.1
Step 3 :Find the area ,using table E.
P(Z>1.1)=
1-P(Z<1.1) =
=0.1375
The probability that household generates more than 30.2 pounds of newspapers is or 31.75%
Note: This PowerPoint is only a summary and your main source should be the book.

43. P(Z<-2.22)= 0.0132 0.0132 × 80 = 1.056 ≈ 1 Example 6-8:
The American Automobile Association reports that the average time it takes to respond to an emergency call is 25 minutes. Assume the variable is approximately normally distributed and the standard deviation is 4.5 minutes. If 80 calls are randomly selected, approximately how many will be responded to in less than 15 minutes?
Solution:
Step 1 : Find the z value .
Step 2 :Draw the figure
-2.22
Step 3 :Find the area ,using table E.
P(Z<-2.22)=
0.0132
Step 4 :to find how many calls .
× 80 = ≈ 1
Note: This PowerPoint is only a summary and your main source should be the book.

44. Finding Data Values Given Specific Probabilities
Formula for Finding X:
we must use that three steps.
First Step: draw the figure and Shading the area.
Second Step: find the z , using table E.
Third Step : find the X ,using the formula .
Note: This PowerPoint is only a summary and your main source should be the book.

45. Example 6-9: Solution: ملاحظة: Top اعلى = bottom اسفل =
To qualify for a police academy, candidates must score in the top 10% on a general abilities test. The test has a mean of 200 and standard deviation of 20. Find the lowest possible score to qualify. Assume the test scores is normally distributed.
Solution:
ملاحظة:
Top اعلى =
bottom اسفل =
Step 1 :Draw the figure
Step 2 : Find the z value . =
Z=1.28
Step 3 :Find the x.
Any body scoring 226 or higher qualify
Note: This PowerPoint is only a summary and your main source should be the book.

48. Example 6-10:
For a medical study, a researcher wishes to select people in the middle 60% of the population based on blood pressure. If the mean systolic blood pressure is 120 and the standard deviation is 8, find the upper and lower readings that would qualify people to participate in the study. Assume that blood pressure readings is normally distributed.
Solution:
Step 1 :Draw the figure
Note: This PowerPoint is only a summary and your main source should be the book.

49. Step 2 : Find the two z values .
=0.8000 =
Step 3 :Find the two x.
The middle 60% will have blood pressure reading of <X<126.72
Note: This PowerPoint is only a summary and your main source should be the book.

50. لمعرفة المطلوب ( Zأو X):
نجد في السؤال المعطيات التالية:
قيمة Z ودائما تكون من 3.4 إلى 3.4-
يوجد إما: the top
أو bottom
أو in the middle
قيمة x ودائما تكون أكبر من 3.4
يوجد إما: less than
أو more than
أو between
طريقة الحل:
لإيجاد X
لإيجاد Z
توجد ثلاث خطوات للحل:
رسم المنحني وتظليل المساحة
إيجاد Z من الجدول
إيجاد X بإستخدام القانون:
إيجاد Z بإستخدام القانون:
إيجاد المساحة من الجدول

51. Determining Normality
Greater than or equal to +1 or
Less than or equal to -1
The data are significantly skewed
The distribution is not normally distributed
Close to zero or equal zero
The data are not significantly skewed
The distribution is normally distributed
Remark
One or two outliers not normally distributed
No outliers normally distributed
Note: This PowerPoint is only a summary and your main source should be the book.

52. إيجاد الوسط الحسابي mean إيجاد الإنحراف المعياري
للتحديد الطبيعي بـ PI
إيجاد الوسط الحسابي mean
إيجاد الوسيط
median
إيجاد الإنحراف المعياري
Standard deviation
التعويض بالقانون
قيمة PI:
إذا كانت محصورة -1≤PI≤+1 :
*التوزيع يتبع التوزيع الطبيعي
إذا كانت قريبة من أوتساوي 0 فإن:
* التوزيع لا يتبع التوزيع الطبيعي
إعداد: أ.أسرار أحمد اليافعي

53. For example:
The data is ( 4 , 7 , 5 , 8 , 12 ) , The standard deviation is Check for normality ?
Solution: MD
Conclude:
The data are not significantly skewed ,so the distribution is approximately normally distributed .
Note: This PowerPoint is only a summary and your main source should be the book.

54. Questions ???
1-If X is normally distributed random variable with µ = 5 , σ = 4 , find the P(x> -1.4) ??

55. Exercises 6-1 page (b)-48(b)-49(c)-50 Exercises 6-2 page (a,b) حلول التمارين الزوجية a) 588 , b)

57. The Central Limit Theorem
6-3
The Central Limit Theorem
نظرية النهاية المركزية
Note: This PowerPoint is only a summary and your main source should be the book.

58. Distribution of Sample Means
توزيع متوسطات العينة
A sampling distribution of sample means
«توزيع المعاينة لمتوسطات العينة»: is a distribution using the means computed from all possible random samples of a specific size taken from a population.
Sampling error» خطأ المعاينة» : is the difference between the sample measure and corresponding population measure due to the fact that the sample is not a perfect representation of the population. ‏
Note: This PowerPoint is only a summary and your main source should be the book.

59. Standard error of the mean
Properties of the distribution of sample means:
1- The mean of the sample means will be the same as the population mean.
2- The standard deviation of the sample means will be smaller than the standard deviation of the population, and it will be equal to the population standard deviation divided by the square root of the sample size.
Standard error of the mean
Note: This PowerPoint is only a summary and your main source should be the book.

60. If the sample size is 40 and the standard deviation of the sample mean
For Example 1 :
If the sample size is 40 and the standard deviation of the sample mean
is 3.95,Then the standard deviation of the population is ????
Solution:
For Example 2 :
The variance of a variable is 16 .If a sample of 80 individuals is selected, Compute the standard error of the mean ?????
Note: This PowerPoint is only a summary and your main source should be the book.

61. The Central Limit Theorem:
As the sample size n increase without limit, the shape of the distribution of the sample means taken with replacement from a population with mean and standard deviation will approach a normal distribution. This distribution will have a mean and a standard deviation
The formula for z values is:
‏ متوسط متوسطات العينة = متوسط المجتمع
الإنحراف المعياري لمتوسط العينة = الإنحراف المعياري للمجتمع ÷ على الجذر التربيعي لحجم العينة
Note: This PowerPoint is only a summary and your main source should be the book.

62. Two important things when we use Central limit theorem
1- when the original variable is normally distributed , the distribution of the sample means will be normally distributed , for any sample size n.
2- when the original variable might not be normally distributed, a sample size must be 30 or more to approximate the distribution of sample means to normal distribution .
To solve the question about sample means we must use three steps.
First Step: find the z , using the formula.
Second Step: draw the figure and Shading the area.
Third Step : find the areas under ,using table E.
Note: This PowerPoint is only a summary and your main source should be the book.

63. Example 6-13: Hours That Children Watch Television
A. C. Neilsen reported that children between the ages of 2 and 5 watch an average of 25 hours of television per week. Assume the variable is normally distributed and the standard deviation is 3 hours. If 20 children between the ages 2 and 5 are randomly selected, find the probability that the mean of the number of hours they watch television will be greater than 26.3 hours?
Solution:
Step 1 : Find the z value .
Note: This PowerPoint is only a summary and your main source should be the book.

64. Step 3 :Find the area ,using table E.
Step 2 :Draw the figure
1.94
Step 3 :Find the area ,using table E.
P(z > 1.94) = – = That is, the probability of obtaining a sample mean greater than is = 2.62%.
Note: This PowerPoint is only a summary and your main source should be the book.

65. Step 1 : Find the two z values .
Example 6-14:
The average age of a vehicle registered in the United States is 8 years, or 96 months. Assume the standard deviation is 16 months. If a random sample of 36 vehicles is selected, find the probability that the mean of their age is between 90 and 100 months.
Solution:
Step 1 : Find the two z values .
Note: This PowerPoint is only a summary and your main source should be the book.

66. Step 3 :Find the area ,using table E.
Step 2 :Draw the figure
-2.25
1.50
Step 3 :Find the area ,using table E.
P(-2.25<Z<1.50)=P(Z<1.50)-P(Z<-2.25) =
= or 92.1 %
The probability of obtaining a sample mean between 90 and 100 months is 92.1%; that is ,P(90 <X<100)= 92.1%
Note: This PowerPoint is only a summary and your main source should be the book.

67. Step 1 : Find the z value . Example 6-15: Meat Consumption
The average number of pounds of meat that a person consumes per year is pounds. Assume that the standard deviation is 25 pounds and the distribution is approximately normal.
Find the probability that a person selected at random consumes less than 224 pounds per year.
If a sample of 40 individuals is selected, find the probability that the mean of the sample will be less than 224 pounds per year.
Solution(a):
Step 1 : Find the z value .
Note: This PowerPoint is only a summary and your main source should be the book.

68. Step 3 :Find the area ,using table E.
Step 2 :Draw the figure
0.22
Step 3 :Find the area ,using table E.
P(Z<0.22) =
0.5871
Note: This PowerPoint is only a summary and your main source should be the book.

69. Solution(b): Step 1 : Find the z value . Step 2 :Draw the figure
1.42
Note: This PowerPoint is only a summary and your main source should be the book.

70. Step 3 :Find the area ,using table E.
P(z < 1.42) =
The probability that mean of a sample of 40 individuals is less than 224 pounds per year is or 92.22%.
Note: This PowerPoint is only a summary and your main source should be the book.

71. EXERCISES 6-3 page Review Exercises page Chapter quiz page