Suppose a population of scores x is normally distributed with μ = 110 and σ = 10. Use the standard normal distribution to find the probability indicated. (Round your answer to four decimal places.)Pr(x < 130) Areas Under the StandardNormal Curve0 Z0The value of A is the area under the standard normal curve between z = 0 and z = z, for z, 2 0. Areas for negative valuesof z, are obtained by symmetry.z.АZ,AZ.AZ.A0.000.00000.430.16640.860.30511.290.40150.010.00400.440.17000.870.30781.300.40320.020.030.040.880.890.00800.450.17360.31061.310.40490.17720.18080.01200.460.31331.320.40660.01600.470.900.31591.330.40820.050.01990.480.18440.910.31861.340.40990.060.02390.490.18790.920.32121.350.41150.070.02790.500.19150.930.32381.360.41310.080.03190.510.19500.940.32641.370.41470.090.03590.520.19850.950.32891.380.41620.03980.04380.100.530.20190.960.33151.390.41770.110.540.20540.970.33401.400.41920.980.33650.42070.42220.120.04780.550.20881.410.130.05170.560.21230.990.33891.420.140.05570.570.21571.000.34131.430.42360.150.05960.580.21901.010.34381.440.42510.06360.06750.160.590.22241.020.34611.450.42651.461.471.481.490.170.600.22571.030.34850.42790.610.620.630.180.07140.22911.040.35080.42920.190.07540.23241.050.35310.43060.200.07930.23571.060.35540.43191.071.080.210.08320.640.23890.35771.500.43320.220.08710.650.24220.35991.510.43450.36211.521.530.230.09100.660.24541.090.43570.240.09480.670.24861.100.36430.43700.250.09870.680.25171.110.36651.540.43820.260.10260.690.25491.120.36861.550.43940.37081.561.570.270.10640.700.25801.130.44060.280.11030.710.26111.140.37290.44180.290.11410.720.26421.150.37491.580.44290.11790.12170.12550.12930.26731.590.44410.44520.44630.300.731.160.37700.37901.601.610.310.740.27041.170.320.750.27341.180.38100.330.760.27641.190.38301.620.44740.44840.44950.45050.45150.45250.45350.45450.340.13310.770.27941.200.38491.630.350.13680.780.28231.210.38691.640.360.14060.790.28521.220.38881.650.370.14430.800.28811.230.39071.660.380.14800.810.29101.240.39251.670.390.15170.820.29391.250.39441.680.400.15540.830.29671.260.39621.690.410.15910.840.29951.270.39801.700.45540.420.16280.850.30231.280.39971.710.4564АР-44

Given that: μ=110, σ=10 Find PX<130.

Answered: Suppose a population of scores x is…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Suppose a population of scores x is normally distributed with

μ = 110

and

σ = 10.

Use the standard normal distribution to find the probability indicated. (Round your answer to four decimal places.)

Pr(x < 130)

Areas Under the Standard
Normal Curve
0 Z0
The value of A is the area under the standard normal curve between z = 0 and z = z, for z, 2 0. Areas for negative values
of z, are obtained by symmetry.
z.
А
Z,
A
Z.
A
Z.
A
0.00
0.0000
0.43
0.1664
0.86
0.3051
1.29
0.4015
0.01
0.0040
0.44
0.1700
0.87
0.3078
1.30
0.4032
0.02
0.03
0.04
0.88
0.89
0.0080
0.45
0.1736
0.3106
1.31
0.4049
0.1772
0.1808
0.0120
0.46
0.3133
1.32
0.4066
0.0160
0.47
0.90
0.3159
1.33
0.4082
0.05
0.0199
0.48
0.1844
0.91
0.3186
1.34
0.4099
0.06
0.0239
0.49
0.1879
0.92
0.3212
1.35
0.4115
0.07
0.0279
0.50
0.1915
0.93
0.3238
1.36
0.4131
0.08
0.0319
0.51
0.1950
0.94
0.3264
1.37
0.4147
0.09
0.0359
0.52
0.1985
0.95
0.3289
1.38
0.4162
0.0398
0.0438
0.10
0.53
0.2019
0.96
0.3315
1.39
0.4177
0.11
0.54
0.2054
0.97
0.3340
1.40
0.4192
0.98
0.3365
0.4207
0.4222
0.12
0.0478
0.55
0.2088
1.41
0.13
0.0517
0.56
0.2123
0.99
0.3389
1.42
0.14
0.0557
0.57
0.2157
1.00
0.3413
1.43
0.4236
0.15
0.0596
0.58
0.2190
1.01
0.3438
1.44
0.4251
0.0636
0.0675
0.16
0.59
0.2224
1.02
0.3461
1.45
0.4265
1.46
1.47
1.48
1.49
0.17
0.60
0.2257
1.03
0.3485
0.4279
0.61
0.62
0.63
0.18
0.0714
0.2291
1.04
0.3508
0.4292
0.19
0.0754
0.2324
1.05
0.3531
0.4306
0.20
0.0793
0.2357
1.06
0.3554
0.4319
1.07
1.08
0.21
0.0832
0.64
0.2389
0.3577
1.50
0.4332
0.22
0.0871
0.65
0.2422
0.3599
1.51
0.4345
0.3621
1.52
1.53
0.23
0.0910
0.66
0.2454
1.09
0.4357
0.24
0.0948
0.67
0.2486
1.10
0.3643
0.4370
0.25
0.0987
0.68
0.2517
1.11
0.3665
1.54
0.4382
0.26
0.1026
0.69
0.2549
1.12
0.3686
1.55
0.4394
0.3708
1.56
1.57
0.27
0.1064
0.70
0.2580
1.13
0.4406
0.28
0.1103
0.71
0.2611
1.14
0.3729
0.4418
0.29
0.1141
0.72
0.2642
1.15
0.3749
1.58
0.4429
0.1179
0.1217
0.1255
0.1293
0.2673
1.59
0.4441
0.4452
0.4463
0.30
0.73
1.16
0.3770
0.3790
1.60
1.61
0.31
0.74
0.2704
1.17
0.32
0.75
0.2734
1.18
0.3810
0.33
0.76
0.2764
1.19
0.3830
1.62
0.4474
0.4484
0.4495
0.4505
0.4515
0.4525
0.4535
0.4545
0.34
0.1331
0.77
0.2794
1.20
0.3849
1.63
0.35
0.1368
0.78
0.2823
1.21
0.3869
1.64
0.36
0.1406
0.79
0.2852
1.22
0.3888
1.65
0.37
0.1443
0.80
0.2881
1.23
0.3907
1.66
0.38
0.1480
0.81
0.2910
1.24
0.3925
1.67
0.39
0.1517
0.82
0.2939
1.25
0.3944
1.68
0.40
0.1554
0.83
0.2967
1.26
0.3962
1.69
0.41
0.1591
0.84
0.2995
1.27
0.3980
1.70
0.4554
0.42
0.1628
0.85
0.3023
1.28
0.3997
1.71
0.4564
АР-44

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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