Question

In: Statistics and Probability

(Typed answers preferable please, no scribbling if hand-written) A sociology student named Adriana is interested in...

(Typed answers preferable please, no scribbling if hand-written)

A sociology student named Adriana is interested in studying the level of danger for police responding to unplanned social protests (i.e. riots). She decided to measure the distance that intoxicated people can throw a beer bottle. (She is also an advocate for beer in cans—they hurt less, at least when they’re empty.) Adriana obtained access to a very large sample of young adults, and asked her study participants to begin by drinking 2 beers. She then had them throw a beer bottle as far as they could. The mean throwing distance was 85 feet, and the standard deviation was 20 feet. She found that throwing distances were normally distributed. Please answer the following questions based on the data.

  1. What is the probability of randomly selecting a young adult who threw 125 feet or more?
  2. If you randomly select a single young adult, what is the probability of selecting someone who threw more than 125 feet or less than 70 feet?
  3. What is the probability randomly selecting two consecutive individuals who threw 60 feet or less?
  4. What two distances capture the middle 95% of all throws?               and                  

Solutions

Expert Solution

mean throwing distance = = 85 feet

standard deviation = = 20 feet

(a) Here if x is the randomly selecting a young adult

P(x > 125 feet) = 1 - P(x < 125 feet)

P(x < 125 feet) = P(x < 125 feet ; 85 feet ; 20 feet)

z = (125 - 85)/20 = 2

Now looking into z table

P(x > 125 feet) = 1 - P(z < 2) = 1 - NORMSDIST(2) = 1 - 0.97725 = 0.02275

(b) Here we have to find

P(70 feet < x < 125 feet) = P(x < 125 feet ; 85 feet ; 20 feet) - P(x < 70 feet ; 85 feet ; 20 feet)

z2 = (125 - 85)/20 = 2

z1 = (70 - 85)/20 = -0.75

P(70 feet < x < 125 feet) = P(Z < 2) - P(Z < -0.75) =NORMSDIST(2)- NORMSDIST(-0.75)

or looking into z table

= 0.97725 - 0.22663 = 0.7506

(c) P(x < 60 feet)

z = (60 - 85)/20 = -1.25

P(x < 60 feet) = NORMSDIST(-1.25) = 0.1056

(d) Here 95% of all throws will capture a distance of +- 1.96   

= 85 +- 1.96 * 20

= (45.8 feet, 124.2 feet)


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