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

Question

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)&#34;-* 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

Accepted Answer

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