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pleaseeeee please help asap, I\'ll be forever grateful Based on a survey with 49

ID: 3326548 • Letter: P

Question

pleaseeeee please help asap, I'll be forever grateful

Based on a survey with 49 patients the Mean time a patient wait at a doctor's office is 65 minutes with standard deviation of 18 minutes. Assuming the waiting time is normally distributed, find the following probabilities. 8. a) In average a patient waits more than 1 hour and 20 minutes? b) In average a patient waits between 70 minutes to 1 hour and 20 minutes? The average time a teenager text a day is 85 times/day with the Standard deviation of 24 times/day. Considering texting time is normally distributed, then construct the 90% confidence intervals) around the true Mean for the given sample sizes below then Interpret 9. a) n=49 b) n=9

Explanation / Answer

Question 8

Number of patients = 49

= 65 minutes

s = 18 minutes

standard error of sample mean se0= 18/49 = 18/7 = 2.5714

Expected sample mean waiting time = 65 minutes

Here if we say that x is the average patient wait time for 49 patients.

(A) Pr(x > 80 minutes) = NORM(x > 80; 65; 2.5714)

Z = (80 - 65)/2.5714 = 5.833

so,

Pr(x > 80 minutes) = Pr(Z > 5.8333) = 1 - Pr(Z < 5.8333) = 1 - 0.99999 = 0.00000 or nearly equal to 0

(b) Pr(70 minutes < x < 80 minutes) = Pr(x < 80 minutes) - Pr(x < 70 minutes) = (Z2) - (Z1)

Z2= (80 - 65)/2.5714 = 5.833

Z1 = (70 - 65)/2.5714 = 1.9444

(Z2) - (Z1) = (5.8333) - (1.9444) = 1 - 0.9741 = 0.0259

Question 9

(a) Here = 85 times/day

= 24 times/day

n = 49

Standard error of sample average time se0= /n = 24/49 = 24/7 = 3.4286

90% confidence interval = +- Z90% se0= 85 +- 1.645 * 3.42586

= (79.36, 90.64)

we can interpret that confidence interval that if we take samples of size 49 repeatedly, 90% chance that the average time a teenager text a day will lie in that confidence interval.

(b) Here iif n = 9

Standard error of sample average time se0= /n = 24/9 = 24/3 = 8

90% confidence interval = +- Z90% se0= 85 +- 1.645 * 8

= (71.84, 98.16)

we can interpret that confidence interval that if we take samples of size 9 repeatedly, 90% chance that the average time a teenager text a day will lie in that confidence interval.