Normal Distributions

Worksheet 4


Exercises

  1. GPAs of SUNY Oswego freshman biology majors have approximately the normal distribution with mean 2.87 and standard deviation .34.
    1. In what range do the middle 90% of all freshman biology majors’ GPAs lie?
    2. Students are thrown out of school if their GPA falls below 2.00. What proportion of all freshman biology majors are thrown out?
    3. What proportion of freshman biology majors have GPA above 3.50?
  2. The length of elephant pregnancies from conception to birth varies according to a distribution that is approximately normal with mean 525 days and standard deviation 32 days.
    1. What percent of pregnancies last more than 600 days (that’s about 20 months)?
    2. What percent of pregnancies last between 510 and 540 days (that’s between 17 and 18 months)?
    3. How short do the shortest 10% of all pregnancies last?
  3. Wingspans of adult herons have approximate normal distribution with mean 125 cm and standard deviation 12 cm.
    1. What proportion of herons have wingspan more than 140 cm?
    2. What is the median wingspan?
    3. What are the first and third quartiles of the wingspans? (25% of all herons have wingspan less than the first quartile; 25% have wingspan more than the third quartile.) Use these to obtain the IQR for heron wingspans.
  4. When robins’ eggs are weighed, it turns out that they vary according to approximately the normal distribution with mean 17.5 mg and standard deviation 3.5 mg.
    1. What proportion of robins’ eggs weigh between 10 mg and 25 mg?
    2. How large are the largest 2% of all robins’ eggs?

Solutions

  1. a) From 2.311 to 3.429; b) 0.0052; c) 0.0322
  2. a) 0.95% (a little less than 1%); b) 36.16%; c) no more than 484 days.
  3. a) 0.1056; b) 125 cm (the mean and median are the same for a normal distribution because of the symmetry); c) the Z-scores for the quartiles are -0.67 and 0.67. That is, quartiles for normal distributions are always 0.67 times the standard deviation below (Q1) and above (Q3) the mean=median. So, for this data, Q1 = 140 - 0.67*12 = 148.0 cm and Q1 = 140 + 0.67*12 = 132.0 cm. Then IQR = 148.0 - 132.0 = 16.0 cm.(For normal distributions IQR = 1.34*s -- always!)
  4. a) 0.9676; b) 24.68 mg.