Question

3-38 PM Wed Oct 9 Use the Lea TUTULT S knewton.com ardes du VIGLUT Question What is the probability that the sample mean for a sample of size 150 will be more than 822 • Use the results from above in your calculation and round your answer to the nearest percent. You may use a calculator or the portion of the z.table given below. Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879 0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549 0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852 0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133 0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389 1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621 1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015 1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633 1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706 1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767 2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.98120.9817 2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.98420.9846 0.9850 0.9854 0.9857 22 0.9861 0.9864 0.9868 0.0871 098750.9878 0.9881 09884 09887 09890

mean of 81 and standard devation of 25

Answer 1

Solution :

Given that ,

mean = $\mu$ = 81

standard deviation = $\sigma$ = 25

n = 150

$\mu$_{$\bar x$} =  $\mu$ = 81

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 25 / $\sqrt$ 150  = 2.0

P($\bar x$ > 82) = 1 - P($\bar x$ < 82)

= 1 - P[($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$} < (82 - 81) / 2.0 ]

= 1 - P(z < 0.50 )

Using z table,    

= 1 - 0.6915

= 0.3085