Question

Assume that a baseball team has an average pitcher, that is, one whose probability of winning any decision is 0.5. If this pitcher has 50 decisions this season, what is the probability that he will win at least 30 games (use a normal curve to calculate this probability).

Question 10 options:

	.0786
	.0835
	.0436
	.1244

Answer 1

Let X be the number of games in pitcher win decisions. The probability of winning any decision is p 0.5. The pitcher has n-50 decisions this season. Thus the exact distribution of X is binomial because X has only two possible outcomes win/not win (Bernoulli trial) and in out of 50 trials we are trying to find the probability of at least 30 wins XBinomial (n-50, p-0.5) In order to find the probability that he will win at least 30 games that is X 2 30. Therefore, the desired probability is PX :30-Σ5030( , ) (0.5)" (1-0.5)50-k This sum is tedious to compute (there are 50-30 + 1 = 21 terms), so we use the normal approximation to the binomial distribution with continuity correction: if np 50*0.5 25 2 10 and n(1-p) 50*(1-0.5) 25 2 10 (which are satisfied in this case), then x-Normal(μ-np, σ-.np(1-p)), approximately Hence to calculate P(X 2 30), we compare the exact distribution of X to a normal distribution with mean μ-np-50*0.5-25 and standard deviation σ-yp (1-p) 50 0.5(1-0.5) 3.5355. 30-25 Z-follows standard normal distribution 3.5355 P(Z> 1.4142) - 1- P(Z< 1.4142) Entry is area under the standard normal curve from-o to z(A) A) ,00 .01 ,02 .03 04 05 06 07 08 09 500504 5080 120 3160 519 5239 5279 5319 5359 5398 5438 5478 5517 5557 5596 5636 5675 5714 5753 5832 5871 5910 5948 5987 6026 6004 6103 6141 6217 6255 6293 633 6368 6406 6443 6480 6517 2579 4653496628 6664 6700 6736 6772 6808 6844 6879 5 6915 6950 685 19 70547088 7123 7157 7190 7224 7910 .7939 7967 95 8023 805 8078 8106 8133 R4388461 8485 8508853 554 8577 8599 862 67257 7 758 8 7881 7291 7611 7324 7357 73897422 74547486 7517 7549 7642 76737704773477647794 7823 7852 981598186.8212 8238 8264 8289 83 8340.8365 8389 1,0 841 1.2884 1.4919 8665 8686 8708 729 8749 8770 8790 8810 8830 88698888 8907 8925 8944 8962 8980 8997 9015 1.1864 9207.9222 9236 925 9265 9279 9345 9357 9370 9382 9394 9406 9418 9429 9441 1.5 933 1.6 945 1.7 95 946394749484 9495 9505 9515 9525 9535 9545 9564 9573 9582 959 99 9608 9616 9625 9633 96499656 9664 9671 9678 9686 9693 9699 9706 97199726 9732 9738 9744 9750 9756 976 9767 1.8 9641 19 971 From standard normal table P(Z<1.41) 0.92017

So, P(X 2 30) 1-0.9207 -0.0793 Since, table is using approximate z 1.41 probability is off by 0.0007, note that; if we use excel function NORM.S.DIST(1.414213562, TRUE) to calculate P(Z<1.4142) we will get the exact probability of P(Z 1.42142) 0.9214 and P(X 2 30) would be 1 0.9214 0.0786 Hence, choice 1 is correct answer