CHAPTER 89 THE NORMAL DISTRIBUTION
CHAPTER 89 THE NORMAL DISTRIBUTION EXERCISE 335 Page 945
1. A component is classed as defective if it has a diameter of less than 69 mm. In a batch of 350 components, the mean diameter is 75 mm and the standard deviation is 2.8 mm. Assuming the diameters are normally distributed, determine how many are likely to be classed as defective.
The z-value corresponding to 69 mm is given by:
x−x
σ
i.e.
69 − 75 = –2.14 standard deviations 2.8
From Table 89.1 on page 944 of the textbook, the area between z = 0 and z = –2.14 is 0.4838, i.e. the shaded area of the diagram below Thus the area to the left of the z = –2.14 ordinate is 0.5000 – 0.4838 = 0.0162 The number likely to be classed as defective = 0.0162 × 350 = 5.67 or 6, correct to nearest whole number
2. The masses of 800 people are normally distributed, having a mean value of 64.7 kg and a standard deviation of 5.4 kg. Find how many people are likely to have masses of less than 54.4 kg.
The z-value corresponding to 54.4 kg is given by:
x−x
σ
i.e.
54.4 − 64.7 = –1.91 standard 5.4
deviations From Table 89.1 on page 944 of the textbook, the area between z = 0 and z = –1.91 is 0.4719, i.e. the shaded area of the diagram below 1378
© 2014, John Bird
Thus the area to the left of the z = –1.91 ordinate is 0.5000 – 0.4719 = 0.0281
The number likely to have masses of less than 54.4 kg = 0.0281 × 800 = 22.48 or 22, correct to nearest whole number
3. 500 tins of paint have a mean content of 1010 ml and the standard deviation of the contents is 8.7 ml. Assuming the volumes of the contents are normally distributed, calculate the number of tins likely to have contents whose volumes are less than (a) 1025 ml, (b) 1000 ml and (c) 995 ml.
(a) The z-value corresponding to 1025 ml is given by:
x−x
σ
i.e.
1025 − 1010 = 1.72 standard 8.7
deviations From Table 89.1 on page 944 of the textbook, the area between z = 0 and z = 1.72 is 0.4573, i.e. the shaded area of the diagram below Thus the area to the left of the z = 1.72 ordinate is 0.5000 + 0.4573 = 0.9573 The number likely to have less than 1025 ml = 0.9573 × 500 = 479, correct to nearest whole number
(b) The z-value corresponding to 1000 ml is given by:
x−x
σ
i.e.
1000 − 1010 = –1.15 standard 8.7
deviations From Table 89.1 on page 944 of the textbook, the area between z = 0 and z = –1.15 is 0.3749, 1379
© 2014, John Bird
i.e. the shaded area of the diagram below Thus the area to the left of the z = –1.15 ordinate is 0.5000 – 0.3749 = 0.1251 The number likely to have less than 1000 ml = 0.1251 × 500 = 63, correct to nearest whole number
(c) The z-value corresponding to 995 ml is given by:
x−x
σ
i.e.
995 − 1010 = –1.72 standard 8.7
deviations From Table 89.1, the area between z = 0 and z = –1.72 is 0.4573, i.e. the shaded area of the diagram below Thus the area to the left of the z = –1.72 ordinate is 0.5000 – 0.4573 = 0.0427 The number likely to have less than 995 ml = 0.0427 × 500 = 21, correct to nearest whole number
4. For the 350 components in Problem 1, if those having a diameter of more than 81.5 mm are rejected, find, correct to the nearest component, the number likely to be rejected due to being oversized.
The z-value corresponding to 81.5 mm is given by:
x−x
1380
σ
i.e.
81.5 − 75 = 2.32 standard deviations 2.8
© 2014, John Bird
From Table 89.1 on page 944 of the textbook, the area between z = 0 and z = 2.32 is 0.4898, i.e. the shaded area of the diagram below Thus the area to the right of the z = 2.32 ordinate is 0.5000 – 0.4838 = 0.0102 The number likely to be classed as oversized = 0.0102 × 350 = 4, correct to nearest whole number
5. For the 800 people in Problem 2, determine how many are likely to have masses of more than (a) 70 kg and (b) 62 kg.
(a) The z-value corresponding to 70 kg is given by:
x−x
σ
i.e.
70 − 64.7 = 0.98 standard 5.4
deviations From Table 89.1 on page 944 of the textbook, the area between z = 0 and z = 0.98 is 0.3365, i.e. the shaded area of the diagram below
Thus the area to the right of the z = 0.98 ordinate is 0.5000 – 0.3365 = 0.1635 The number likely to have masses of more than 70 kg = 0.1635 × 800 = 130.8 or 131, correct to nearest whole number (b) The z-value corresponding to 62 kg is given by:
x−x
σ
i.e.
62 − 64.7 = –0.50 standard 5.4
deviations 1381
© 2014, John Bird
From Table 89.1 on page 944 of the textbook, the area between z = 0 and z = –0.50 is 0.1915, i.e. the shaded area of the diagram below
Thus the area to the right of the z = –0.50 ordinate is 0.5000 + 0.1915 = 0.6915 The number likely to have masses of more than 62 kg = 0.6915 × 800 = 553.2 or 553, correct to nearest whole number
6. The mean diameter of holes produced by a drilling machine bit is 4.05 mm and the standard deviation of the diameters is 0.0028 mm. For 20 holes drilled using this machine, determine, correct to the nearest whole number, how many are likely to have diameters of between (a) 4.048 and 4.0553 mm and (b) 4.052 and 4.056 mm, assuming the diameters are normally distributed.
(a) The z-value corresponding to 4.048 mm is given by:
x−x
σ
i.e.
4.048 − 4.05 = –0.71 standard 0.0028
deviations From Table 89.1 on page 944 of the textbook, the area between z = 0 and z = –0.71 is 02611 The z-value corresponding to 4.0553 mm is given by:
x−x
σ
i.e.
4.0553 − 4.05 = 1.89 standard 0.0028
deviations From Table 89.1, the area between z = 0 and z = 1.89 is 0.4706 The probability of the diameter being between 4.048 mm and 4.0553 mm is 0.2611 + 0.4706 = 0.7317 (see shaded area in diagram below) The number likely to have a diameter between 4.048 mm and 4.0553 mm = 0.7317 × 20 = 14.63 = 15, correct to nearest whole number
1382
© 2014, John Bird
(b) The z-value corresponding to 4.052 mm is given by:
x−x
σ
i.e.
4.052 − 4.05 = 0.71 standard 0.0028
deviations From Table 89.1, the area between z = 0 and z = 0.71 is 0.2611 The z-value corresponding to 4.056 mm is given by:
x−x
σ
i.e.
4.056 − 4.05 = 2.14 standard 0.0028
deviations From Table 89.1, the area between z = 0 and z = 2.14 is 0.4838 The probability of the diameter being between 4.052 mm and 4.056 mm is 0.4838 – 0.2611 = 0.2227 (see shaded area in diagram below) The number likely to have a diameter between 4.052 mm and 4.056 mm = 0.2227 × 20 = 4.454 = 4, correct to nearest whole number
7. The intelligence quotients of 400 children have a mean value of 100 and a standard deviation of 14. Assuming that IQs are normally distributed, determine the number of children likely to have IQs of between (a) 80 and 90, (b) 90 and 110 and (c) 110 and 130
(a) The z-value corresponding to an IQ of 80 is given by:
x−x
σ
i.e.
80 − 100 = –1.43 standard 14
deviations From Table 89.1 on page 944 of the textbook, the area between z = 0 and z = –1.43 is 0.4236
1383
© 2014, John Bird
The z-value corresponding to an IQ of 90 is given by:
x−x
σ
i.e.
90 − 100 = –0.71 standard 14
deviations From Table 89.1, the area between z = 0 and z = –0.71 is 0.2611 The probability of IQ being between 80 and 90 is 0.4236 – 0.2611 = 0.1625 (see shaded area in diagram below)
The number likely to have an IQ between 80 and 90 = 0.1625 × 400 = 65 (b) The z-value corresponding to an IQ of 110 is given by:
x−x
σ
i.e.
110 − 100 = 0.71 standard 14
deviations From Table 89.1, the area between z = 0 and z = 0.71 is 0.2611 The probability of IQ being between 90 and 110 is 0.2611 + 0.2611 = 0.5222 (see shaded area in diagram below)
The number likely to have an IQ between 90 and 110 = 0.5222 × 400 = 208.88 = 209, correct to nearest whole number (c) The z-value corresponding to an IQ of 130 is given by:
x−x
σ
i.e.
130 − 100 = 2.14 standard 14
deviations From Table 89.1, the area between z = 0 and z = 2.14 is 0.4838 The probability of IQ being between 110 and 130 is 0.4838 – 0.2611 = 0.2227 (see shaded 1384
© 2014, John Bird
area in diagram below)
The number likely to have an IQ between 110 and 130 = 0.2227 × 400 = 89.08 = 89, correct to nearest whole number
8. The mean mass of active material in tablets produced by a manufacturer is 5.00 g and the standard deviation of the masses is 0.036 g. In a bottle containing 100 tablets, find how many tablets are likely to have masses of (a) between 4.88 and 4.92 g, (b) between 4.92 and 5.04 g and (c) more than 5.04 g.
(a) The z-value corresponding to 4.88 g is given by:
x−x
σ
i.e.
4.88 − 5.00 = –3.33 standard 0.036
deviations From Table 89.1 on page 944 of the textbook, the area between z = 0 and z = –3.33 is 0.4996. The z-value corresponding to 4.92 g is given by:
4.92 − 5.00 = –2.22 standard deviations. 0.036
From Table 89.1, the area between z = 0 and z = –2.22 is 0.4868 The probability of having masses between 4.88 g and 4.92 g is 0.4996 – 0.4868 = 0.0128 (see shaded area in diagram below) The number of tablets likely to have a mass between 4.88 g and 4.92 g = 0.0128 × 100 = 1, correct to nearest whole number
1385
© 2014, John Bird
(b) The z-value corresponding to 4.92 g is –2.22 standard deviations, from above, and the area between z = 0 and z = –2.22 is 0.4868 The z-value corresponding to 5.04 g is given by:
x−x
σ
i.e.
5.04 − 5.00 = 1.11 standard 0.036
deviations From Table 89.1, the area between z = 0 and z = 1.11 is 0.3665 The probability of having masses between 4.92 g and 5.04 g is 0.4868 + 0.3665 = 0.8533 (see shaded area in diagram below) The number of tablets likely to have a mass between 4.92 g and 5.04 g = 0.8533 × 100 = 85, correct to nearest whole number
(c) The z-value corresponding to 5.04 g is 1.11 standard deviations, from above, and the area between z = 0 and z = 1.11 is 0.3665 The probability of having a mass greater than 5.04 g is 0.5000 – 0.3665 = 0.1335 (see shaded area in diagram below) The number of tablets likely to have a mass greater than 5.04 g = 0.1335 × 100 = 13, correct to nearest whole number
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© 2014, John Bird
EXERCISE 336 Page 948
1. A frequency distribution of 150 measurements is as shown: Class mid-point value
26.4 26.6 26.8 27.0 27.2 27.4 27.6
Frequency
5
12
24
36
36
25
12
Use normal probability paper to show that this data approximates to a normal distribution and hence determine the approximate values of the mean and standard deviation of the distribution. Use the formula for mean and standard deviation to verify the results obtained.
To test the normality of a distribution, the upper class boundary values are plotted against percentage cumulative frequency values on normal probability paper The table below shows the upper class boundary values for the distribution, together with the cumulative frequency and percentage cumulative frequency
Class mid-point
Upper class
value
boundary
Frequency
Cumulative
% cumulative
frequency
frequency
value 26.4
26.5
5
5
5/150 = 3
26.6
26.7
12
5 + 12 = 17
17/150 = 11
26.8
26.9
24
17 + 24 = 41
41/150 = 27
27.0
27.1
36
77
51
27.2
27.3
36
113
75
27.4
27.5
25
138
92
27.6
27.7
12
150
100
The coordinates of upper class boundary values/percentage cumulative frequency values are shown plotted below. Since the points plotted lie very nearly in a straight line, the data is approximately normally distributed From the graph, the mean occurs at 50%, i.e. mean, x = 27.1 at point P At 84% cumulative frequency value, i.e. point Q, upper class boundary value = 27.38 At 16% cumulative frequency value, i.e. point R, upper class boundary value = 26.78 © 2014, John Bird 1387
Hence, standard deviation, σ =
27.38 − 26.78 0.6 = 0.3 = 2 2
By calculation, mean,
x =
=
( 5 × 26.4 ) + (12 × 26.6 ) + ( 24 × 26.8) + ( 36 × 27.0 ) + ( 36 × 27.2 ) + ( 25 × 27.4 ) + (12 × 27.6 ) 150 4061.8 = 27.079 150
5 ( 26.4 − 27.079 )2 + 12 ( 26.6 − 27.079 )2 + 24 ( 26.8 − 27.079 )2 + ... 150
Standard deviation, σ =
=
13.51175 = 0.3001 150 1388
© 2014, John Bird
2. A frequency distribution of the class mid-point values of the breaking loads for 275 similar fibres is as shown below: Load (kN)
17 19 21
23
25
27
29
31
Frequency
9
78
64
28
14
4
23 55
Use normal probability paper to show that this distribution is approximately normally distributed and determine the mean and standard deviation of the distribution (a) from the graph and (b) by calculation.
To test the normality of a distribution, the upper class boundary values are plotted against percentage cumulative frequency values on normal probability paper The table below shows the upper class boundary values for the distribution, together with the cumulative frequency and percentage cumulative frequency Class mid-point
Upper class
value (kN)
boundary
Frequency
Cumulative
% cumulative
frequency
frequency
value 17
18
9
9
9/275 = 3
19
20
23
9 + 23 = 32
32/275 = 12
21
22
55
32 + 55 = 87
87/275 = 32
23
24
78
165
60
25
26
64
229
83
27
28
28
257
93
29
30
14
271
99
31
32
4
275
100
The coordinates of upper class boundary values/percentage cumulative frequency values are shown plotted below. Since the points plotted lie very nearly in a straight line, the data is approximately normally distributed (a) From the graph, the mean occurs at 50%, at point P , i.e. mean, x = 23.5 kN At 84% cumulative frequency value, i.e. point Q, upper class boundary value = 26.2 At 16% cumulative frequency value, i.e. point R, upper class boundary value = 20.4 © 2014, John Bird 1389
Hence, standard deviation, = σ
(b) By calculation, mean, x =
=
26.2 − 20.4 5.8 = 2.9 kN = 2 2
( 9 ×17 ) + ( 23 ×19 ) + ( 55 × 21) + ( 78 × 23) + ( 64 × 25) + ...... 275 6425 = 23.364 kN 275
9 (17 − 23.364 )2 + 23 (19 − 23.364 )2 + 55 ( 21 − 23.364 )2 + .... 275
Standard deviation, σ =
=
2339.6364 = 2.917 kN 275
1390
© 2014, John Bird
1. A component is classed as defective if it has a diameter of less than 69 mm. In a batch of 350 components, the mean diameter is 75 mm and the standard deviation is 2.8 mm. Assuming the diameters are normally distributed, determine how many are likely to be classed as defective.
The z-value corresponding to 69 mm is given by:
x−x
σ
i.e.
69 − 75 = –2.14 standard deviations 2.8
From Table 89.1 on page 944 of the textbook, the area between z = 0 and z = –2.14 is 0.4838, i.e. the shaded area of the diagram below Thus the area to the left of the z = –2.14 ordinate is 0.5000 – 0.4838 = 0.0162 The number likely to be classed as defective = 0.0162 × 350 = 5.67 or 6, correct to nearest whole number
2. The masses of 800 people are normally distributed, having a mean value of 64.7 kg and a standard deviation of 5.4 kg. Find how many people are likely to have masses of less than 54.4 kg.
The z-value corresponding to 54.4 kg is given by:
x−x
σ
i.e.
54.4 − 64.7 = –1.91 standard 5.4
deviations From Table 89.1 on page 944 of the textbook, the area between z = 0 and z = –1.91 is 0.4719, i.e. the shaded area of the diagram below 1378
© 2014, John Bird
Thus the area to the left of the z = –1.91 ordinate is 0.5000 – 0.4719 = 0.0281
The number likely to have masses of less than 54.4 kg = 0.0281 × 800 = 22.48 or 22, correct to nearest whole number
3. 500 tins of paint have a mean content of 1010 ml and the standard deviation of the contents is 8.7 ml. Assuming the volumes of the contents are normally distributed, calculate the number of tins likely to have contents whose volumes are less than (a) 1025 ml, (b) 1000 ml and (c) 995 ml.
(a) The z-value corresponding to 1025 ml is given by:
x−x
σ
i.e.
1025 − 1010 = 1.72 standard 8.7
deviations From Table 89.1 on page 944 of the textbook, the area between z = 0 and z = 1.72 is 0.4573, i.e. the shaded area of the diagram below Thus the area to the left of the z = 1.72 ordinate is 0.5000 + 0.4573 = 0.9573 The number likely to have less than 1025 ml = 0.9573 × 500 = 479, correct to nearest whole number
(b) The z-value corresponding to 1000 ml is given by:
x−x
σ
i.e.
1000 − 1010 = –1.15 standard 8.7
deviations From Table 89.1 on page 944 of the textbook, the area between z = 0 and z = –1.15 is 0.3749, 1379
© 2014, John Bird
i.e. the shaded area of the diagram below Thus the area to the left of the z = –1.15 ordinate is 0.5000 – 0.3749 = 0.1251 The number likely to have less than 1000 ml = 0.1251 × 500 = 63, correct to nearest whole number
(c) The z-value corresponding to 995 ml is given by:
x−x
σ
i.e.
995 − 1010 = –1.72 standard 8.7
deviations From Table 89.1, the area between z = 0 and z = –1.72 is 0.4573, i.e. the shaded area of the diagram below Thus the area to the left of the z = –1.72 ordinate is 0.5000 – 0.4573 = 0.0427 The number likely to have less than 995 ml = 0.0427 × 500 = 21, correct to nearest whole number
4. For the 350 components in Problem 1, if those having a diameter of more than 81.5 mm are rejected, find, correct to the nearest component, the number likely to be rejected due to being oversized.
The z-value corresponding to 81.5 mm is given by:
x−x
1380
σ
i.e.
81.5 − 75 = 2.32 standard deviations 2.8
© 2014, John Bird
From Table 89.1 on page 944 of the textbook, the area between z = 0 and z = 2.32 is 0.4898, i.e. the shaded area of the diagram below Thus the area to the right of the z = 2.32 ordinate is 0.5000 – 0.4838 = 0.0102 The number likely to be classed as oversized = 0.0102 × 350 = 4, correct to nearest whole number
5. For the 800 people in Problem 2, determine how many are likely to have masses of more than (a) 70 kg and (b) 62 kg.
(a) The z-value corresponding to 70 kg is given by:
x−x
σ
i.e.
70 − 64.7 = 0.98 standard 5.4
deviations From Table 89.1 on page 944 of the textbook, the area between z = 0 and z = 0.98 is 0.3365, i.e. the shaded area of the diagram below
Thus the area to the right of the z = 0.98 ordinate is 0.5000 – 0.3365 = 0.1635 The number likely to have masses of more than 70 kg = 0.1635 × 800 = 130.8 or 131, correct to nearest whole number (b) The z-value corresponding to 62 kg is given by:
x−x
σ
i.e.
62 − 64.7 = –0.50 standard 5.4
deviations 1381
© 2014, John Bird
From Table 89.1 on page 944 of the textbook, the area between z = 0 and z = –0.50 is 0.1915, i.e. the shaded area of the diagram below
Thus the area to the right of the z = –0.50 ordinate is 0.5000 + 0.1915 = 0.6915 The number likely to have masses of more than 62 kg = 0.6915 × 800 = 553.2 or 553, correct to nearest whole number
6. The mean diameter of holes produced by a drilling machine bit is 4.05 mm and the standard deviation of the diameters is 0.0028 mm. For 20 holes drilled using this machine, determine, correct to the nearest whole number, how many are likely to have diameters of between (a) 4.048 and 4.0553 mm and (b) 4.052 and 4.056 mm, assuming the diameters are normally distributed.
(a) The z-value corresponding to 4.048 mm is given by:
x−x
σ
i.e.
4.048 − 4.05 = –0.71 standard 0.0028
deviations From Table 89.1 on page 944 of the textbook, the area between z = 0 and z = –0.71 is 02611 The z-value corresponding to 4.0553 mm is given by:
x−x
σ
i.e.
4.0553 − 4.05 = 1.89 standard 0.0028
deviations From Table 89.1, the area between z = 0 and z = 1.89 is 0.4706 The probability of the diameter being between 4.048 mm and 4.0553 mm is 0.2611 + 0.4706 = 0.7317 (see shaded area in diagram below) The number likely to have a diameter between 4.048 mm and 4.0553 mm = 0.7317 × 20 = 14.63 = 15, correct to nearest whole number
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© 2014, John Bird
(b) The z-value corresponding to 4.052 mm is given by:
x−x
σ
i.e.
4.052 − 4.05 = 0.71 standard 0.0028
deviations From Table 89.1, the area between z = 0 and z = 0.71 is 0.2611 The z-value corresponding to 4.056 mm is given by:
x−x
σ
i.e.
4.056 − 4.05 = 2.14 standard 0.0028
deviations From Table 89.1, the area between z = 0 and z = 2.14 is 0.4838 The probability of the diameter being between 4.052 mm and 4.056 mm is 0.4838 – 0.2611 = 0.2227 (see shaded area in diagram below) The number likely to have a diameter between 4.052 mm and 4.056 mm = 0.2227 × 20 = 4.454 = 4, correct to nearest whole number
7. The intelligence quotients of 400 children have a mean value of 100 and a standard deviation of 14. Assuming that IQs are normally distributed, determine the number of children likely to have IQs of between (a) 80 and 90, (b) 90 and 110 and (c) 110 and 130
(a) The z-value corresponding to an IQ of 80 is given by:
x−x
σ
i.e.
80 − 100 = –1.43 standard 14
deviations From Table 89.1 on page 944 of the textbook, the area between z = 0 and z = –1.43 is 0.4236
1383
© 2014, John Bird
The z-value corresponding to an IQ of 90 is given by:
x−x
σ
i.e.
90 − 100 = –0.71 standard 14
deviations From Table 89.1, the area between z = 0 and z = –0.71 is 0.2611 The probability of IQ being between 80 and 90 is 0.4236 – 0.2611 = 0.1625 (see shaded area in diagram below)
The number likely to have an IQ between 80 and 90 = 0.1625 × 400 = 65 (b) The z-value corresponding to an IQ of 110 is given by:
x−x
σ
i.e.
110 − 100 = 0.71 standard 14
deviations From Table 89.1, the area between z = 0 and z = 0.71 is 0.2611 The probability of IQ being between 90 and 110 is 0.2611 + 0.2611 = 0.5222 (see shaded area in diagram below)
The number likely to have an IQ between 90 and 110 = 0.5222 × 400 = 208.88 = 209, correct to nearest whole number (c) The z-value corresponding to an IQ of 130 is given by:
x−x
σ
i.e.
130 − 100 = 2.14 standard 14
deviations From Table 89.1, the area between z = 0 and z = 2.14 is 0.4838 The probability of IQ being between 110 and 130 is 0.4838 – 0.2611 = 0.2227 (see shaded 1384
© 2014, John Bird
area in diagram below)
The number likely to have an IQ between 110 and 130 = 0.2227 × 400 = 89.08 = 89, correct to nearest whole number
8. The mean mass of active material in tablets produced by a manufacturer is 5.00 g and the standard deviation of the masses is 0.036 g. In a bottle containing 100 tablets, find how many tablets are likely to have masses of (a) between 4.88 and 4.92 g, (b) between 4.92 and 5.04 g and (c) more than 5.04 g.
(a) The z-value corresponding to 4.88 g is given by:
x−x
σ
i.e.
4.88 − 5.00 = –3.33 standard 0.036
deviations From Table 89.1 on page 944 of the textbook, the area between z = 0 and z = –3.33 is 0.4996. The z-value corresponding to 4.92 g is given by:
4.92 − 5.00 = –2.22 standard deviations. 0.036
From Table 89.1, the area between z = 0 and z = –2.22 is 0.4868 The probability of having masses between 4.88 g and 4.92 g is 0.4996 – 0.4868 = 0.0128 (see shaded area in diagram below) The number of tablets likely to have a mass between 4.88 g and 4.92 g = 0.0128 × 100 = 1, correct to nearest whole number
1385
© 2014, John Bird
(b) The z-value corresponding to 4.92 g is –2.22 standard deviations, from above, and the area between z = 0 and z = –2.22 is 0.4868 The z-value corresponding to 5.04 g is given by:
x−x
σ
i.e.
5.04 − 5.00 = 1.11 standard 0.036
deviations From Table 89.1, the area between z = 0 and z = 1.11 is 0.3665 The probability of having masses between 4.92 g and 5.04 g is 0.4868 + 0.3665 = 0.8533 (see shaded area in diagram below) The number of tablets likely to have a mass between 4.92 g and 5.04 g = 0.8533 × 100 = 85, correct to nearest whole number
(c) The z-value corresponding to 5.04 g is 1.11 standard deviations, from above, and the area between z = 0 and z = 1.11 is 0.3665 The probability of having a mass greater than 5.04 g is 0.5000 – 0.3665 = 0.1335 (see shaded area in diagram below) The number of tablets likely to have a mass greater than 5.04 g = 0.1335 × 100 = 13, correct to nearest whole number
1386
© 2014, John Bird
EXERCISE 336 Page 948
1. A frequency distribution of 150 measurements is as shown: Class mid-point value
26.4 26.6 26.8 27.0 27.2 27.4 27.6
Frequency
5
12
24
36
36
25
12
Use normal probability paper to show that this data approximates to a normal distribution and hence determine the approximate values of the mean and standard deviation of the distribution. Use the formula for mean and standard deviation to verify the results obtained.
To test the normality of a distribution, the upper class boundary values are plotted against percentage cumulative frequency values on normal probability paper The table below shows the upper class boundary values for the distribution, together with the cumulative frequency and percentage cumulative frequency
Class mid-point
Upper class
value
boundary
Frequency
Cumulative
% cumulative
frequency
frequency
value 26.4
26.5
5
5
5/150 = 3
26.6
26.7
12
5 + 12 = 17
17/150 = 11
26.8
26.9
24
17 + 24 = 41
41/150 = 27
27.0
27.1
36
77
51
27.2
27.3
36
113
75
27.4
27.5
25
138
92
27.6
27.7
12
150
100
The coordinates of upper class boundary values/percentage cumulative frequency values are shown plotted below. Since the points plotted lie very nearly in a straight line, the data is approximately normally distributed From the graph, the mean occurs at 50%, i.e. mean, x = 27.1 at point P At 84% cumulative frequency value, i.e. point Q, upper class boundary value = 27.38 At 16% cumulative frequency value, i.e. point R, upper class boundary value = 26.78 © 2014, John Bird 1387
Hence, standard deviation, σ =
27.38 − 26.78 0.6 = 0.3 = 2 2
By calculation, mean,
x =
=
( 5 × 26.4 ) + (12 × 26.6 ) + ( 24 × 26.8) + ( 36 × 27.0 ) + ( 36 × 27.2 ) + ( 25 × 27.4 ) + (12 × 27.6 ) 150 4061.8 = 27.079 150
5 ( 26.4 − 27.079 )2 + 12 ( 26.6 − 27.079 )2 + 24 ( 26.8 − 27.079 )2 + ... 150
Standard deviation, σ =
=
13.51175 = 0.3001 150 1388
© 2014, John Bird
2. A frequency distribution of the class mid-point values of the breaking loads for 275 similar fibres is as shown below: Load (kN)
17 19 21
23
25
27
29
31
Frequency
9
78
64
28
14
4
23 55
Use normal probability paper to show that this distribution is approximately normally distributed and determine the mean and standard deviation of the distribution (a) from the graph and (b) by calculation.
To test the normality of a distribution, the upper class boundary values are plotted against percentage cumulative frequency values on normal probability paper The table below shows the upper class boundary values for the distribution, together with the cumulative frequency and percentage cumulative frequency Class mid-point
Upper class
value (kN)
boundary
Frequency
Cumulative
% cumulative
frequency
frequency
value 17
18
9
9
9/275 = 3
19
20
23
9 + 23 = 32
32/275 = 12
21
22
55
32 + 55 = 87
87/275 = 32
23
24
78
165
60
25
26
64
229
83
27
28
28
257
93
29
30
14
271
99
31
32
4
275
100
The coordinates of upper class boundary values/percentage cumulative frequency values are shown plotted below. Since the points plotted lie very nearly in a straight line, the data is approximately normally distributed (a) From the graph, the mean occurs at 50%, at point P , i.e. mean, x = 23.5 kN At 84% cumulative frequency value, i.e. point Q, upper class boundary value = 26.2 At 16% cumulative frequency value, i.e. point R, upper class boundary value = 20.4 © 2014, John Bird 1389
Hence, standard deviation, = σ
(b) By calculation, mean, x =
=
26.2 − 20.4 5.8 = 2.9 kN = 2 2
( 9 ×17 ) + ( 23 ×19 ) + ( 55 × 21) + ( 78 × 23) + ( 64 × 25) + ...... 275 6425 = 23.364 kN 275
9 (17 − 23.364 )2 + 23 (19 − 23.364 )2 + 55 ( 21 − 23.364 )2 + .... 275
Standard deviation, σ =
=
2339.6364 = 2.917 kN 275
1390
© 2014, John Bird
