Question

The price of shares of Bank of Florida at the end of trading each day for the last year followed the normal distribution. Assume there were 240 trading days in the year. The mean price was $42.00 per share and the standard deviation was $2.25 per share. (Round your answers to 2 decimal places.)

a-1. What is the probability that the end-of-day trading price is over $45.00? ______%

Answer 1

Solution :

    we are given that : The price of shares of Bank of Florida at the end of trading each day for the last year followed the normal distribution.

   Assume there were 240 trading days in the year. The mean price was $42.00 per share and the standard deviation was $2.25 per share.

Thus   $Mean = \mu =42.00$    and Standard Deviation = $\sigma =2.25$

We have to find :

Part A. What is the probability that the end of day trading price is over $45.00?

That is we have to find : P( X > 45.00) = ..............?

Thus find z score for x = 45.00

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

$z = \frac{45.00-42.00 }{2.25 }$

$z = \frac{3.00 }{2.25 }$

$z = 1.33$

Thus we get :

P( X > 45.00 ) = P( Z > 1.33)

P( X > 45.00 ) = 1 - P( Z < 1.33)

Look in z table for z = 1.3 and 0.03 and find corresponding area.

01 02 03 05 07 0.0 .5000 .5040 5080 .520 5160 5199 5239 5279 .5319 5359 0.1.5398 5438 5478 0.2 .5793 .5832 .5871 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 6517 0.4 .6554 .6591 .6628 .φ64 .6700 .6736 .6772 .6808 .6844 .6879 0.5 6915 6950 .6985 0.6 7257 .7291 .7324 0.7 .7580 0.8 .7881 .7910 7939 767 .7995 8023 8051 8078 .8106 8133 0.9 .8159 8186 8212 .8138 .8264 .8289 8315 8340 8365 8389 1.0 8413 8438 8461 885 8508 8531 8554 8577 8599 .8621 1.1 8643 8665 .8686 1.2 8849 8869 8888 .8907 .8925 .8944 8962 8980 8997 9015 17 5557 5596 5636 5675.5714 5753 10 .5948 5987 .6026 6064 6103 6141 19 .7054 7088 7123 7157 57 7389 7422 .7454 7486 .7517 7549 573 7704 7734 7764 7794 7823 7611 7642 .8729 8749 .8770 8790 .8810 8830 0 9032 onaam -90 66一.9082 .9099 .9115 .9131 .9147 .9162 .9177 1.4 .9192 9207 9222 9236 9251 9265 9279 9292 9306 .9319

Thus we get ; P( Z < 1.33 ) = 0.9082

Thus P( X > 45.00 ) = 1 - P( Z < 1.33)

        P( X > 45.00 ) = 1 - 0.9082

       P( X > 45.00 ) = 0.0918