According to Reader's Digest, 38% of primary care doctors think their patients receive unnecessary medical care. a. Suppose a sample of 350 primary care doctors was taken. Show the sampling distribution of the proportion of the doctors who think their patients receive unnecessary medical care. Use z-table. E(p)=p=(to4decimals) b. What is the probability that the sample proportion will be within 0.03 of the population proportion. Round your answer to four decimals. c. What is the probability that the sample proportion will be within 0.05 of the population proportion. Round your answer to four decimals. d. What would be the effect of taking a larger sample on the probabilities in parts (b) and (c)? Why? The probabilities would . This is because the increase in the sample size makes the standard error, p, TABLE 1 CUMULATIVE PROBABILITIES FOR THE STANDARD NORMAL DISTRIBUTION (Continued) Entries in the table give the area under the curve to the left of the z value. For example, for z=1.25, the cumulative probability is .8944. According to Reader's Digest, 38% of primary care doctors think their patients receive unnecessary medical care. a. Suppose a sample of 350 primary care doctors was taken. Show the sampling distribution of the proportion of the doctors who think their patients receive unnecessary medical care. Use z-table. E(p)=p=(to4decimals) b. What is the probability that the sample proportion will be within 0.03 of the population proportion. Round your answer to four decimals. c. What is the probability that the sample proportion will be within 0.05 of the population proportion. Round your answer to four decimals. d. What would be the effect of taking a larger sample on the probabilities in parts (b) and (c)? Why? The probabilities would . This is because the increase in the sample size makes the standard error, p, TABLE 1 CUMULATIVE PROBABILITIES FOR THE STANDARD NORMAL DISTRIBUTION (Continued) Entries in the table give the area under the curve to the left of the z value. For example, for z=1.25, the cumulative probability is .8944