In: Statistics and Probability
Problem FIVE: Form B
The distribution of weights for 18 year old high school males in the Toledo area can be approximated by a Normal Curve with Mean weight of 173 pounds and Standard Deviation of 15.0 pounds.
a…..What is the probability that a randomly selected student is somewhere between 170 and 175 pounds?
b…..What is the probability that a randomly selected sample of 100 students had a mean weight between 170 and 175 pounds?
c…..What weight would put a student in the 35th percentile?
ANSWERS
a)
b)
c)
Problem SIX:
Short Answer…….
__________1) If the probability of event A occurring is 0.63, what is the probability that the Complement of event A occurs?
__________2) If the Standard Deviation for a study was 1.96, what was its Variance?
__________3) What z-score corresponds to the 85%-ile?
__________4) For a Left-Tailed t-test, what would be the Critical Value if 13 people participated in this study?
Set α = 0.05
__________5) How many random samples of seven different names can be drawn from a population that contains a total of 26 names?
_________ 6) 78% of all cases in a Normal Population fall between what +/ - z-scores?
__________7) Your score of 90 on a calculus class gave you a z-score of +2.00. If the class Mean was 80.00, what was the Standard Deviation?
Problem 5
a)
Given,
= 173, = 15
We convert this to standard normal as
P( X < x) = P( Z < x - / )
So,
P( 170 < X < 175) = P( X < 175) - P( X < 170)
= P( Z < 175 - 173 / 15) - P( Z < 170 - 173 / 15)
= P( Z < 0.1333) - P( Z < -0.2)
= 0.5530 - 0.4207
= 0.1323
b)
Using central limit theorem,
P( < x) = P( Z < x - / / sqrt(n) )
So,
P( 170 < < 175) = P( < 175) - P( < 170)
= P( Z < 175 - 173 / 15 / sqrt(100) ) - P( Z < 170 - 173 / 15 / sqrt(100) )
= P( Z < 1.3333) - P( Z < -2)
= 0.9088 - 0.0228
= 0.8860
c)
pth percentile = + Z * , Where Z is critical value at pth confidence level.
So,
35th percentile = 173 + -0.3853 * 15
= 167.22