(a)&nbsp;(i) According to a survey, one in five Americans claim to have seen a doctor during the past month. If 10 people are selected at random, find the binomial probability that exactly 4 will have visited a doctor last month.&nbsp;(ii) An average sales company receives 4 calls per hour on their toll-free line. For any given hour, find the probability that it will receive less than equal 3 calls.&nbsp;(b) Full-time Ph.D students receive an average of $12,837 per year. If the average salaries are normally distributed with a standard deviation of $1500, find these probabilities.&nbsp;(i) The student makes more than $14,000.&nbsp;(ii) The student makes between $13,000 and $15,000.

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(a)&amp;nbsp;(i) According to a survey, one in five Americans claim to have seen a doctor during the past month. If 10 people are selected at random, find the binomial probability that exactly 4 will have visited a doctor last month.&amp;nbsp;(ii) An average sales company receives 4 calls per hour on their toll-free line. For any given hour, find the probability that it will receive less than equal 3 calls.&amp;nbsp;(b) Full-time Ph.D students receive an average of $12,837 per year. If the average salaries are normally distributed with a standard deviation of $1500, find these probabilities.&amp;nbsp;(i) The student makes more than $14,000.&amp;nbsp;(ii) The student makes between $13,000 and $15,000.

Camem1937 · Accepted Answer

Step 1 A. i. n=10 p=0.20 X=number of people who visit doctor X follows binomial distribution with parameters n and p P(X=4)=0.088080 ii. X= number of phone calls X follows Poisson distribution with parameter 4 P(X≤3)=P(X=0)+P(X=1)+P(X=2)+P(X=3) =0.0183+0.0733+0.1465+0.1954 =0.4335 Step 2 b. X= salaries Average of X=$12837 Standard deviation of X=s=$1500 i. P(X&amp;gt;14000) Using standard normal distribution Z, =P(Z&amp;gt;0.775333) =0.2191 ii. P(13000&amp;lt;X&amp;lt;15000) =P(0.1086667&amp;lt;Z&amp;lt;1.442) =0.0141

(a) (i) A survey found that one out of five Americans say he or sh

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