Elena, a 15-year-old girl, is 58 inches tall. Based on what we know, the average height for
Question:
Elena, a 15-year-old girl, is 58 inches tall. Based on what we know, the average height for girls at this age is 63.80 inches, with a standard deviation of 2.66 inches.
a. Calculate Elena’s z score.
b. What percentage of girls are taller than Elena?
c. What percentage of girls are shorter?
d. How much would Elena have to grow to be perfectly average?
e. If Sarah is in the 75th percentile for height at age 15, how tall is she?
f. How much would Elena have to grow in order to be at the 75th percentile with Sarah?
Fantastic news! We've Found the answer you've been seeking!
Step by Step Answer:
a b 4854 of the data fall between this z score and the mean Because this is a ne...View the full answer
Answered By
Shadrack Mulunga
I am a Biochemistry by profession. However, I have explored different fields of study. My quest to explore new fields has helped me gain new knowledge and skills in Business, clinical psychology, sociology, organizational behavior and general management, and Project Management. I count my expertise in Project management, in particular, creation of Work Break Down Structure (WBS) and use of Microsoft Project software as one of my greatest achievement in Freelancing industry. I have helped thousands of BSC and MSC students to complete their projects on time and cost-effectively using the MS Project tool. Generally, I find happiness in translating my knowledge and expertise to success of my clients. So far, i have helped thousands of students to not only complete their projects in time but also receive high grades in their respective courses. Quality and timely delivery are the two key aspects that define my work. All those who hired my services always come back for my service. If you hire my services today, you will surely return for more. Try me today!
154+ Reviews
289+ Question Solved
Related Book For 
Essentials Of Statistics For The Behavioral Sciences
ISBN: 9781464107771
3rd Edition
Authors: Susan A. Nolan
Question Posted:
