please only answer D, E and F!!! ONLY D, E AND F!! d) Using this portion of the standard normal chart and part (c), find the probability P(x 2 60)..00.01.022.03.04.059656.972697839671.9738.97931.8.964196781.92.0.9713.9772.9649.9719.9778966497329788.9744.9798e) You should have found that the answer to (d) is much smaller than the answer to (b). This means thatthe probability of getting 60% or more tails depends on the sample size n. That is:Probability of 6 or more tails out of 10> Probability of 60 or more tails out of 100even though the proportion, p = 60%, is the same in both cases. Why does this happen?%3Dstrelive fechan shove YoAll youd inf) Without doing any new calculations, resolve these two hypothesis tests to a 3% level of significance.The null hypothesis is that the coin is fair.The alternate hypothesis is that the coin is biased toward tails.Họ:p = 0.5H:p> 0.5I. If the coin is flipped 10 times and it comes up tails 6 times, what is your conclusion? Compare yourresult from part (b) to the cutoff of 3%.II. If the coin is flipped 100 times and it comes up tails 60 times, what is your conclusion? Compareyour result from part (d) to the cutoff of 3%. 27. This problem is about probabilities and coin flipping. The goal is to compare the probability of gettingtails or greater when flipping a coin 10 times, versus the probability of getting 60 tails or greater whenflipping a coin 100 times. In each case, the proportion of tails is the same, 2 60%, but the probabilities aredifferent. Assume that we have a fair coin, so the odds of getting tails on any one flip are 0.5.a) First, let's calculate the odds of getting exactly 6 tails out of 10 total coin flips. Use the binomialprobability formula P(x) = „Cz (p)*(1-p)"-* to find P(6).%3DCAp(X=6)=10= 0. 205078b) You don't have to find the other probabilities by hand; they can be found from this piece10 0.0011.0102.0443 .1174 2055 24620550of the binomial chart. Using this information, determine the probability of getting 6 tails ormore. That is, find P(x 2 6). Hint: it could be 6 or 7 or 8 or 9 or 10. Note that the answer topart (a) is right there next to x = 6, so you can check to see if you got part (a) correct.P(X 26): P (x 6) + PlX=7) TP (x =8)*P(x=9) +P(x-lo)= 0.205 t0.117+0. 044+0.07o Fo.00 =0.37692%3D67 .1178.0449.01010 .001Now we would like to consider n = 100 coin flips and determine P(x 2 60). It is too difficult to add upeach of the possibilities by hand from the binomial chart, so instead we will approximate the binomialdistribution as a normal distribution. This is valid because n is large enough.c) Using the appropriate z-score formula, determine the z-score for ôộ =60= 0.60, p = .5, and n = 100.100- o.6-0.5Vo.5 *(i-0.5)2=2Vo.0025100

Given information:- PTail=12=0.50PHead=12=0.50

d) Using this portion of the standard normal chart and part (c), find the probability P(x 2 60). 00 .01 .02 .03 .04 .05 9678 .9744 .9798 1.8 .9641 .9713 9772 9649 1.9 2.0 9719 9778 .9656 .9726 9783 9664 .9732 9788 9671 9738 9793 e) You should have found that the answer to (d) is much smaller than the answer to (b). This means that the probability of getting 60% or more tails depends on the sample size n. That is: Probability of 6 or more tails out of 10> Probability of 60 or more tails out of 100 even though the proportion, p = 60%, is the same in both cases. Why does this happen? %3D ive e You Allyou in f) Without doing any new calculations, resolve these two hypothesis tests to a 3% level of significance. The null hypothesis is that the coin is fair. The alternate hypothesis is that the coin is biased toward tails. Ho:p = 0.5 H:p > 0.5 %3D I. If the coin is flipped 10 times and it comes up tails 6 times, what is your conclusion? Compare your result from part (b) to the cutoff of 3%. II. If the coin is flipped 100 times and it comes up tails 60 times, what is your conclusion? Compare your result from part (d) to the cutoff of 3%.