Question

In: Statistics and Probability

Problem FIVE:                                        &n

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?

Solutions

Expert Solution

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

orchestra answered 2 years ago
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