Suppose a manufacturer of Vaporizer cartridges samples 200 cartridges and measures a failure rate of 1.5% (3 cartridges). Using the Normal approximation to the binomial, construct a 95% confidence interval for the probability of being defective. Table A.1 (continued) Binomial Probability Sums b(r; n, p)n p 0.100.200.250.3019 0 0.1351 0.0144 0.0042 0.001110.40 0.500.00010.4203 0.0829 0.0310 0.0104 0.0008 0.000020.7054 0.236930.8850 0.455140.9648 0.67335 0.9914 0.836960.998370.999781.0000910111213141516171819T=0P171819200.600.700.800.900.11130.0462 0.0055 0.0004 0.00000.26310.1332 0.0230 0.00220.00010.46540.0006 0.00000.00310.00010.0116 0.00060.0352 0.0028 0.00000.2822 0.0696 0.00960.6678 0.4739 0.1629 0.03180.9324 0.8251 0.6655 0.3081 0.08350.9767 0.9225 0.8180 0.4878 0.17960.9933 0.9713 0.9161 0.6675 0.3238 0.0885 0.0105 0.00030.9984 0.9911 0.9674 0.8139 0.5000 0.1861 0.0326 0.00160.9997 0.99770.98950.9115 0.6762 0.3325 0.0839 0.0067 0.00001.0000 0.9995 0.9972 0.9648 0.8204 0.5122 0.1820 0.0233 0.00030.9165 0.6919 0.3345 0.0676 0.00170.9682 0.8371 0.5261 0.1631 0.00860.9994 0.9904 0.9304 0.7178 0.3267 0.03520.9999 0.9978 0.9770 0.8668 0.5449 0.11501.0000 0.9996 0.9945 0.9538 0.7631 0.29460.99920.57970.9999 0.9989 0.98560.9999 0.9994 0.98841.0000 0.9999 0.99691.00001.00000.9896 0.91710.86491.0000 1.0000 1.00001.000020 00.12160.01150.00320.00080.000010.39170.06920.02430.00760.00050.000020.0355 0.00360.000230.0013 0.00000.1071 0.01600.2375 0.05100.4164 0.12560.0059 0.00030.00000.0207 0.00160.057760.0065 0.00035 0.9887 0.80420.9976 0.91337 0.9996 0.96798 0.9999 0.99001.00000.1316 0.0210 0.0013 0.00000.0565 0.00510.000190.6769 0.2061 0.09130.8670 0.4114 0.22524 0.9568 0.6296 0.41480.61720.7858 0.6080 0.25000.8982 0.7723 0.41590.9591 0.8867 0.5956 0.25170.9974 0.9861 0.9520 0.7553 0.41190.9994 0.9961 0.9829 0.8725 0.5881 0.2447 0.0480 0.00260.00000.9999 0.9991 0.9949 0.9435 0.7483 0.4044 0.1133 0.0100 0.00011.0000 0.9998 0.9987 0.9790 0.8684 0.58410.9997 0.9935 0.9423 0.75001.0000 0.9984 0.9793 0.8744 0.5836 0.19580.9997 0.9941 0.9490 0.7625 0.37041.0000 0.9987 0.9840 0.8929 0.58860.1275 0.01710.00061011120.2277 0.03210.0004131.00000.3920 0.08670.0024140.0113150.0432160.13300.9998 0.9964 0.9645 0.7939 0.32311.0000 0.9995 0.9924 0.9308 0.60831.0000 0.9992 0.9885 0.87841.0000 1.0000 1.0000 Table A.1 (continued) Binomial Probability Sums Σb(x; n, p)z=00.300.0075 0.00230.0501 0.019320.0774 0.01230.00120.0001n r 0.10 0.20 0.2517 0 0.1668 0.02251 0.4818 0.11820.7618 0.3096 0.16373 0.9174 0.5489 0.3530 0.2019 0.0464 0.0064 0.0005 0.00004 0.9779 0.7582 0.5739 0.3887 0.1260 0.0245 0.0025 0.00015 0.9953 0.8943 0.7653 0.5968 0.2639 0.0717 0.0106 0.0007 0.00006 0.9992 0.9623 0.8929 0.7752 0.4478 0.1662 0.0348 0.00320.00010.9999 0.9891 0.9598 0.8954 0.6405 0.3145 0.0919 0.0127 0.00051.0000 0.9974 0.9876 0.9597 0.8011 0.5000 0.1989 0.0403 0.0026 0.00000.9995 0.9969 0.9873 0.9081 0.6855 0.3595 0.1046 0.0109 0.00010.9999 0.9994 0.9968 0.9652 0.8338 0.5522 0.2248 0.03771.0000 0.9999 0.9993 0.9894 0.9283 0.73611.0000 0.9999 0.9975 0.9755 0.87401.0000 0.9995 0.9936 0.95360.9999 0.9988 0.98771.0000789100.0008110.4032 0.10570.0047120.6113 0.24180.02211314150.7981 0.4511 0.08260.9226 0.6904 0.23820.9999 0.9979 0.9807 0.8818 0.51821.0000 0.9998 0.9977 0.9775 0.83321.0000 1.0000 1.0000 1.0000161718012345678910111213156181417P0.15010.0180 0.00560.45030.09910.03950.7338 0.27130.13530.9018 0.5010 0.30570.9718 0.71640.9936 0.86710.9988 0.94870.99981.00000.00160.00010.00000.0142 0.00130.00010.0600 0.0082 0.00070.1646 0.0328 0.00380.5187 0.3327 0.0942 0.01540.7175 0.5344 0.20880.8610 0.7217 0.37430.8593 0.56340.9404 0.73680.9790 0.86530.9424 0.75970.9998 0.9986 0.97971.0000 0.99971.00000.9837 0.94310.9957 0.98070.9991 0.99460.9998 0.9988 0.99391.00000.400.500.0002 0.00000.0021 0.0001 0.00000.600.700.800.00000.00020.0013 0.00000.0481 0.0058 0.00030.1189 0.0203 0.00140.2403 0.0576 0.00610.4073 0.1347 0.02100.00000.00020.00090.5927 0.2632 0.0596 0.00430.4366 0.1407 0.01630.2783 0.05130.4656 0.13290.900.00000.00020.8811 0.62570.00120.9942 0.9519 0.79120.00640.9987 0.9846 0.90580.6673 0.28360.02820.99980.9962 0.96720.8354 0.4990 0.09821.00000.9993 0.9918 0.9400 0.7287 0.26620.9999 0.9987 0.9858 0.9009 0.54971.0000 0.99991.00000.9984 0.9820 0.84991.00001.0000 1.0000

Suppose a manufacturer of Vaporizer cartridges samples 200 cartridges and measures a failure rate of 1.5% (3 cartridges). Using the Normal approximation to the binomial, construct a 95% confidence interval for the probability of being defective.

Suppose a manufacturer of Vaporizer cartridges samples 200 cartridges and measures a failure rate of 1.5% (3 cartridges). Using the Normal approximation to the binomial, construct a 95% confidence interval for the probability of being defective.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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