Question

Normal distribution

The execution times in seconds of a certain experiment have a normal distribution with a mean of 38 and a variance of 36.
a) What is the probability that an experiment test will last more than 42 seconds?
b) If the test times located in the top 10% are classified as reserved, what is the minimum time to be in this category?
c) What is the proportion of tests that have times that exceed at least five seconds at 25% of the shorter times?

0,0 0,1 1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1,0 1.1 1.2 1,3 1.4 1.5 1.6 1,7 1.8 1.9 0,5000 0,5398 0.5793 0.6179 0.6554 0.6915 0.7257 0.7580 0,7881 0,8159 0,8413 0.8643 0,8849 0,9032 0,9192 09337 0.9452 0.9554 0,9641 0.9713 0.9772 0.9821 0.9861 0.9893 0.9918 0,9938 0.9953 0.9965 0,9974 0.9981 0.9987 0.9990 0.9993 0.9995 0.9997 0.9998 0.9998 0.9999 0.9999 1,0000 0.5040 0,5438 0.5832 0,6217 0.6591 0.6950 0.7291 0.7611 0.7910 0.8186 0.8438 0.8665 0,8869 0.9049 0.9207 0.9345 0.9463 0.9564 0.9649 0.9719 0.9778 0,9826 0.9864 0.9896 0.9920 0.9940 0.9955 0.9966 0.9975 0.9982 0.9987 0.9991 0,9993 0.9995 0.9997 0.9998 0.9998 0.9999 0.9999 1,0000 2.0 0,5080 0,5478 0,5871 0.6255 0,6628 0,6985 0.7324 0.7642 0,7939 0.8212 0,8461 0,8686 0.8888 0.9066 0,9222 0.9357 0,9474 0,9573 0,9656 0,9726 0,9783 0.9830 0.9868 0.9898 0.9922 0,9941 0.9956 0.9967 0,9976 0.9982 0.9987 0,9991 0.9994 0,9995 0,9997 0.9998 0,9999 0.9999 0.9999 1,0000 0,5120 0,5517 0.5910 0.6293 0,6664 0.7019 0.7357 0.7673 0.7967 0,8238 0.8485 0,8708 0,8907 0.9082 0.9236 0.9370 09494 0.9582 0.9664 0.9732 0.9788 0.9834 0.9871 0.9901 0.9925 0.9943 0.9957 0.9968 0.9977 0.9983 0.9988 0.9991 0.9994 0.9996 0.9997 0.9998 0.9999 0.9999 0.9999 1,0000 0,5160 0,5557 0.5948 0,6331 0,6700 0.7054 0.7389 0.7704 0,7995 0,8264 0,8508 0,8729 0,8925 0,9099 0,9251 0.9382 0.9495 0.9591 0.9671 0.9738 0.9793 0.9838 0.9875 0,9904 0,9927 0,9945 0.9959 0.9969 0,9977 0,9984 0,9988 0,9992 0.9994 0,9996 0,9997 0.9998 0.9999 0,9999 0.9999 1,0000 0.5199 0,5596 0,5987 0.6368 0,6736 0.7088 07422 0.7734 0.8023 0.8289 0.8531 0.8749 0.8944 0.9115 0.9265 0.9394 0.9505 0.9599 0.9678 09744 0.9798 0.9842 0.9878 0.9906 0.9929 0.9946 0.9960 0.9970 0.9978 0.9984 0.9989 0,9992 0.9994 0,9996 0.9997 0.999.8 0 9999 0.9999 0.9999 1,0000 7 0,5279 0.5675 0,6064 0,6443 0.6808 0.7157 0.7486 0.7794 0.8078 0.8340 0.8577 0,8790 0,8980 0.9147 0.9292 09418 09525 0.9616 0.9693 0.9756 0.9808 0.9850 0.9884 0.9911 0.9932 0.9949 0.9962 0.9972 0.9979 0.9985 0.9989 0,9992 0.9995 0,9996 0.9997 0.9998 0.9999 0.9999 0.9999 1,0000 0,5239 0,5636 0.6026 0,6406 0,6772 0.7123 0.7454 0,7764 0,8051 0.8315 0.8554 0,8770 0,8962 0,9131 0,9279 0.9406 0.9515 0.9.609 0.9686 0.9750 0.9.803 0,9846 0,9881 0,9909 0,9931 0.9948 0.9961 0.9971 0.9979 0.9985 0.9989 0,9992 0.9994 0.9996 0.9997 0,9998 0.9999 0.9999 0.9999 1,0000 0,5319 0,5714 1 0,6103 0,6480 0.6844 0,7190 07517 0,7823 0,8106 0,8365 0.8599 0,8810 10,8997 0,9162 0,9306 0429 0.9535 0.9625 0,9699 0.9761 0.9.812 0.9854 0.9887 0,9913 0.9934 0,9951 1 0.9963 0.9973 0.9980 0.9986 0,9990 0,9993 0.9995 0,9996 0.9997 0.9998 0.9999 0.9999 0.9999 1,0000 0,5359 0.5753 0.6141 0.6517 0.6879 0.7224 0.7549 0.7852 0.8133 0.8389 0.8621 0,8830 0.9015 0.9177 0.9319 0.9441 0.9545 0.9633 0.9706 0.9767 0.9817 0.9857 0.9890 0.9916 0.9936 0.9952 0.9964 0.9974 0.9981 0.9986 0.9990 0,9993 0.9995 0.9997 0.9998 0.9998 0.9999 0.9999 0.9999 1,0000 2,1 2,2 2,3 2,4 2,5 2.6 2,7 2,8 2.9 3,0 3,2 3,3 3.4 3,5 3.6 3.7 3.8 3,9

Answer 1

Solution:

Given, the Normal distribution with,

  $\mu$ = 38

$\sigma$² = 36

$\sigma$ = 6

a) Find P(X > 42 seconds )

P(X > 42) = P[(X - $\mu$ )/$\sigma$ >  (42 - $\mu$ )/$\sigma$]

= P[Z > (42 - 38)/6]

= P[Z > 0.67]

= 1 - P[Z < 0.67]

=    1 - 0.7486 ( use z table)

=    0.2514

b) Let x be the required value.

For top 10% data,

P(X > x) = 10% = 0.10

Therefore, P(X < x) = 0.90

P[(X - $\mu$ )/$\sigma$ <  (x - $\mu$ )/$\sigma$] = 0.90

P[Z <  (x - 38)/6] = 0.90

But from z table , 0.90 probability for Z < 1.28

P[Z < 1.28 ] = 0.90

$\therefore$  (x - 38)/6 = 1.28

$\therefore$ x =  45.68

The minimum time to be in top 10% category is  45.68

c)First we find P(exceed at least five seconds )  i.e. P(X $\geq$ 5)  

P(X  $\geq$  5) = P[(X - $\mu$ )/$\sigma$ >  (5 - $\mu$ )/$\sigma$]

= P[Z > (5 - 38)/6]

= P[Z > -5.5]

= 1 - P[Z < -5.50]

=    1 - 0 ( use z table)

=    1.0000

Required prop0rtion = 1.0000 * 0.25 = 0.25