Question

IN EXCEL SO I CAN SEE THE FORMULAS USED.

Open the Excel program Suppose we want to know the standard normal distribution up to the given value. That is P(Z <a). Lets use Excel instead of the standard normal distribution table In cell A1 type mean. In cell B1 type 0. That means that we are letting mean be zero. click on cell A2 and type SD. In cell B2 type 1. That means that we are letting standard deviation be one. That is, the normal distribution is standard. Suppose that a=-1.72. Click on cell B6 and type =NORMDIST(-1.72.B1,B2,TRUE). We obtained in B6 the probability of Z<-1.72. Compare this result with what you would find in the standard normal distribution table. Now we do it for any normal distribution. That is, not only that we do not need the table, but we also do not have to do the algebra to go from standard normal distribution to the given normal distribution. We will illustrate that on an example. Suppose that SAT scores have normal distribution with mean 840 and standard deviation 200. If you score 970 on the SAT, what is your percentile? To do this we first go to cell D1and type 840. That means that we are letting mean be 840. Click on cell D2 and type 200. That means that we are letting standard deviation be 200. That is, the normal distribution is not standard. Now click on cell D6 and type =NORMDIST(970, D1, D2,TRUE). Did you obtain approximately 74 as your percentile? Exercises

Answer 1

Exercise 1 :

B A 1 Mean 2 SD 3 4 Plz<1.34236) 0 1 =NORMDIST(1.34236,B1,B2,TRUE) 0.91026

P( z < 1.34236 ) = 0.91026

Exercise 2 :

To find table value T(1.34236 ) we need to round the number 1.34236 to 2 decimal places

So we have to find the table value for 1.34 , therefore we find corresponding value in the row 1.3 and across the column 0.04

Z .00 01 .02 03 .04 .05 1.06 .07 .08 .19 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 12 13 1.4 15 -5000 .5040 .5080 .5120 5160 519 15199 .5239 15279 5319 15319 5359 5398 1.54 38 5478 .5517 .5557 .5596 5636 156755714 5753 1.5793 1,5832 .5871 .5910 5948 5987,6026 16064 .6103 .6141 16179 .6217 .6255 .6293 .6331 .6368 .6406 6443.6480 .6480 .6517 1.6554 .6591.6628 .6664 6700 .6736 6772 .6772.6808 1.6808 .6844 .6844 ,6879 1.6915 .6950 .6985 .7019 7054 7088 7123 7157 7190 7190 7224 7257 7291 7291 7324 7357 7389 1.7422 7454 1.7486 .7517 7517 7549 7580 7611 1.7611.7642.7673 7704 7734 7764 .7794 .7823 78237852 17881 7910 7939 79677995 .8023.8051 .8078 .8106 8133 1.8159 8186 1.8186 8212 8212 18238.8264 1829 18315 18340 8365 1839 18413 84 38 18488 8461,8485 8508 .8531 .8554 8577.8599 18621 8643 1.8665 8686 8708 8729 1.8749 8770 18790 8810 1.8810 830 1,8849,8869 8888 ,8907.8925 1.8944.8962 .8980 8997 015 .9032 ,9049 9066 49082 9099 9115 9131 19147 9162 9162 9177 9192 9207 192079222 9222.9236 19236 9251 9265 ,9279 19292 9306 9306 9319 19332 19345 9357 19370 9382 19394 9406 19418 9429 19441

So table valueT(1.34) is 0.9099

Exercise 3: The values are different but approximately equal ,the excel gives the accurate value for 1.34236 but to use table we have to round it to two decimal places ,so it is approximated value .

Exercise 4 :

A B с 1 Mean 890 2 SD | 150 3 4 P(x<1048) =NORMDIST(1048,B1,B2,TRUE) 10.853906

P( x < 1048 ) = 0.853906