The mean clotting time of blood is 7.45 seconds with a standard deviation of 3.6

Question

The mean clotting time of blood is 7.45 seconds with a standard deviation of 3.6 seconds. What is the probability that an individual's clotting time will be less than 4

seconds or greater than 11 seconds? Assume a normal distribution. This problem can be solved using technology or the table for the standard normal curve from your text.

Area under a normal curve to the left of z, where x = StartFraction x - mu Over sigma EndFraction

0.4761

Area under a normal curve to the left of z, where x = StartFraction x - mu Over sigma EndFraction z 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 -3.4 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0002 -3.3 0.0005 0.0005 0.0005 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0003 -3.2 0.0007 0.0007 0.0006 0.0006 0.0006 0.0006 0.0006 0.0005 0.0005 0.0005 -3.1 0.001 0.0009 0.0009 0.0009 0.0008 0.0008 0.0008 0.0008 0.0007 0.0007 -3 0.0013 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.0011 0.001 0.001 -2.9 0.0019 0.0018 0.0018 0.0017 0.0016 0.0016 0.0015 0.0015 0.0014 0.0014 -2.8 0.0026 0.0025 0.0024 0.0023 0.0023 0.0022 0.0021 0.0021 0.002 0.0019 -2.7 0.0035 0.0034 0.0033 0.0032 0.0031 0.003 0.0029 0.0028 0.0027 0.0026 -2.6 0.0047 0.0045 0.0044 0.0043 0.0041 0.004 0.0039 0.0038 0.0037 0.0036 -2.5 0.0062 0.006 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051 0.0049 0.0048 -2.4 0.0082 0.008 0.0078 0.0075 0.0073 0.0071 0.0069 0.0068 0.0066 0.0064 -2.3 0.0107 0.0104 0.0102 0.0099 0.0096 0.0094 0.0091 0.0089 0.0087 0.0084 -2.2 0.0139 0.0136 0.0132 0.0129 0.0125 0.0122 0.0119 0.0116 0.0113 0.011 -2.1 0.0179 0.0174 0.017 0.0166 0.0162 0.0158 0.0154 0.015 0.0146 0.0143 -2 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183 -1.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.025 0.0244 0.0239 0.0233 -1.8 0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.0294 -1.7 0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 0.0375 0.0367 -1.6 0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0.0475 0.0465 0.0455 -1.5 0.0668 0.0655 0.0643 0.063 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559 -1.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0721 0.0708 0.0694 0.0681 -1.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823 -1.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.102 0.1003 0.0985 -1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.123 0.121 0.119 0.117 -1 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379 -0.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.166 0.1635 0.1611 -0.8 0.2119 0.209 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.1867 -0.7 0.242 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177 0.2148 -0.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483 0.2451 -0.5 0.3085 0.305 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.281 0.2776 -0.4 0.3446 0.3409 0.3372 0.3336 0.33 0.3264 0.3228 0.3192 0.3156 0.3121 -0.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557 0.352 0.3483 -0.2 0.4207 0.4168 0.4129 0.409 0.4052 0.4013 0.3974 0.3936 0.3897 0.3859 -0.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.4247 negative .0 0.5 0.496 0.492 0.488 0.484 0.4801

0.4761

0.4721 0.4681 0.4641 The mean clotting time of blood is 7.45 seconds with a standard deviation of 3.6 seconds. What is the probability that an individual's clotting time will be less than 4 seconds or greater than 11 seconds? Assume a normal distribution. This problem can be solved using technology or the table for the standard normal curve from your text Click here to view page 1 of the table. Click here to view page 2 of the table The probability is (Round to four decimal places as needed.)

Explanation / Answer

Solution:

The probability that an individual's clotting time will be less than 4 seconds or greater than 11 seconds

P(x < 4) + P(x > 11) = ?

z = (4 - 7.45)/3.6 = -0.9583

z = (11 - 7.45)/3.6 = 0.9861


P(x < 4) + P(x > 11) = P(z < -0.9583) + P(z > 0.9861)

P ( Z<0.9861 ) can be found by using the following standard normal table

We see that P ( Z<0.9861 )=0.8389.

We see that P ( Z<0.9583 )=0.8315 so,

P ( Z<0.9583)=1P ( Z<0.9583 )=10.8315=0.1685

At the end we have:

P (0.9583<Z<0.9861 )=0.6704

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