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DOUAhharks People Window Help Secur https://www. Do Homework Kimberly Pacheco aspx?hor meworkid-410259608& Math 112: 168flushed true&cld; 4320003&centerw.; TTh 1-3 Homework: Section 6.2 Homework Score: 0 of 1 pt e 6.2.46 16 of 21 (7 complete) According to HW Score: 55.16%, 11.58 of 21 pts Normally the data, the mean quantitative score on standardized female high school seniors was 500. scores are approximately Click distributed with a population standard deviation A scholarship committee wants to give awards to women who score Click here to view on the test. What score does an applicant need? Include a well-labeled Normal curve as part of your answer here to page 1of the view page 2 of the Standard Normal Table An applicant would need a score of at least to be in the 99th percentile or above. (Round to the nearest integer as needed.) Enter your answer in the answer box and then click Check Answer. 1 part remaining Clear AI Question 21 (0.6771 X Question 20 (0/1) V Question 19 (1/1)

Explanation / Answer

Problem 6.2.46

We are given

Mean = 500

SD = 50

Critical Z value for 99th percentile = 2.326348

X = Mean + Z*SD

X = 500 + 2.326348*50 = 616.3174

Required answer = 616

Problem 6.2.20.a

We have to find area to the left of a z-score of -0.39

P(Z<-0.39) = 0.348268

Required probability = 0.3483

(By using excel)

Problem 6.2.20.b

Here, we have to find the area to the right of z-score of -0.39

P(Z>-0.39) = 1 – P(Z<-0.39) = 1 – 0.3483 = 0.6517

Required probability = 0.6517

(By using excel)

Problem 6.2.34

We are given

Mean = 515, SD = 50

We have to find P(X<600)

Z = (X – mean) / SD

Z = (600 – 515) / 50 = 85/50 = 1.7

P(X<600) = P(Z<1.7) = 0.955435

Required percentage = 95.5%

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