Question

Assume that a random variable is normally distributed with a mean of 1,200 and a variance of 360. Complete parts a through c below. What is the probability that a randomly selected value will be greater than 1, 253​?

Answer 1

Solution:
Given: a random variable is normally distributed with a mean of 1,200 and a variance of 360.

Thus X ~ Normal $(\mu =1200 \: , \: \sigma = 360)$

Find: the probability that a randomly selected value will be greater than 1, 253

P( X> 1253) = .............?

Find z score for x = 1253

$z=\frac{x-\mu }{\sigma }$

$z=\frac{1253-1200 }{360}=\frac{53 }{360}=0.15$

Thus we get:

P( X> 1253) = P( Z > 0.15)

P( X> 1253) = 1 - P( Z < 0.15)

Look in z table for z = 0.1 and 0.05 and find corresponding area.

.00 .01 .02 .03 .04 .05 .06 .07 .08 .09 Z 5 99 5359 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 .5040 .5080 5120 .5160 5239 5279 .5319 .5000 5478 5675 .5714 5753 5390 E428 5517 5557 5596 .5636 6141 .6517 5793 .5832 5871 5910 5948 .5987 6026 6064 6103 6368 6406 .6443 6480 6179 .6217 .6255 6628 .6293 .6331 .6844 6879 7224 7549 6554 6591 6664 6700 6736 6772 6808 7019 .6950 .6985 7054 7088 7123 7157 7190 .6915 7486 7517 7257 7291 7324 7357 7389 7422 7454 7734 7580 7611 7642 7673 7704 7764 7794 7823 7852 7967 7995 8023 8051 8078 .8106 .8133 7881 7910 7939 .8238 8159 .8186 .8438 .8212 .8264 8289 8315 8340 8365 .8389 8554 1.0 1.1 1.2 .8461 8485 .8508 8531 8577 .8599 .8621 .8413 .8686 .8729 8790 .8810 .8830 8643 8849 .8665 8869 8708 8749 8770 8888 8907 8925 8944 .8962 8980 .8997 .9015

P( Z< 0.15) = 0.5596

Thus

P( X> 1253) = 1 - P( Z < 0.15)

P( X> 1253) = 1 - 0.5596

P( X> 1253) = 0.4404