Question

The following data (excluded outlier) is collected from English Fall session homework #4. Assume that distribution is normally distributed.

sample size 32 ,mean 64 ,Sample standard deviation 15.82 ,population standard deviation 12.5 ,max 90 ,Q1 60 ,Median 69.5, Q3 75

Your homework #4 Score is 32

1. Analyze your score based on the date above.

2. Assume that distribution is bell-shaped, find the percentage who scored greater than your homework score. Analyze your homework score based on the percentage you found.

3. Find 90% confidence interval for the population mean. (a) Interpret the result. (b) How do you analyze your homework score from (a)?

4. For English , the passing rate is 53%. Find the probability that at least 23 students pass your course.

5. If a English department chair says that the average homework 4 score ( entire students) for English is 60/100 pts. Do you believe it? Show your work.

Show which function if you use for TI84 please.

Answer 1

Angwer: Analysis of the score 32 z-score of 32 32-64--2.56 Za 12-5 Here the score is a standard deviations below The mean Hence, the performance of the student is extremely low P(x>32) *100% =P (**04 > 2004) 100% = - L - P (2 L-Ls62] oo. L-06-2-56)] 100% = {1-0.00523361) 100% - 0.994-76639 X100% = 49.48% c) The 90% confidence interval for (* 9 x Toos! to XT.CS) Given mean 64 6 = 12.5 n=32 To.os = 1,645

t is believable to accept that (64.- (1695); 64 + C1.645); 64 +125 (1,645)) = (60-3650 292, 67.6349 708) - 660-37 , 67:63) Interpretation with 90% confidence we conclude that the mean score of the student is contained in The interval (60.37,67-63) The score does not belong to the confidence interward, it is outside the c I of the mean. d) Binomial distribution P(x= x)=(13).601533 (1–0:53)*** P(x223) = 3 (34) (0:53)}-o-szjeg Ho: M=60 H: M 60 in (7-60) To Test statistic 132 (64-60) 12.5 = 1.81019336 Here it is observed that (T) =1-810 2T0.67 -960 So The mean score is so X=23 20023386 367