I am having some trouble with this problem. Could someone help me out? 3. [20 po

Question

I am having some trouble with this problem. Could someone help me out?

3. [20 points] Suppose that the weights of people who work in an office building are normally distributed with a mean of -165 lb and a standard deviation of = 25 lb. [4 points each] (a) What is the probability that one person, selected at random from the building, weighs (b) Suppose that three people are selected independently of each other. Find the probability (c) Find the probability that a sample of three people will have a mean weight greater than (d) What is the conceptual difference between parts (b) and (c), i.e. how are the events (e) Have you ever ridden in an elevator and read a sign stating its maximum load and more than 200 lb? that all three weigh more than 200 lb, i.e. find P(X, > 200n X2 > 200n X, > 200) 200 lb different? wondered about the chances the elevator would be overloaded?What is the chance that the total weight of five people is more than 1000 lb, i.e. what is P(X,t...t Xs > 1000)? Hint try to re-express this as an X-style problem.

Explanation / Answer

3.

NORMAL DISTRIBUTION

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2

standard normal distribution is a normal distribution with a,

mean of 0,

standard deviation of 1

equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)

mean ( u ) = 165

standard Deviation ( sd )= 25

a.

P(X > 200) = (200-165)/25

= 35/25 = 1.4

= P ( Z >1.4) From Standard Normal Table

= 0.0808

b.

3 people are selected independently of each other,p(A B)= p(A).p(B)

p(X1>200 X2>200 X3>200) = p(X1>200).P(X2>200).p(X3>200)

= 0.0808*0.0808*0.0808 = 0.00052

c.

NORMAL DISTRIBUTION

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2

standard normal distribution is a normal distribution with a,

mean of 0,

standard deviation of 1

equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)

mean ( u ) = 165

standard Deviation ( sd )= 25/ Sqrt ( 3 ) =14.4338

sample size (n) = 3

P(X > 200) = (200-165)/25/ Sqrt ( 3 )

= 35/14.434= 2.4249

= P ( Z >2.4249) From Standard Normal Table

= 0.0077

d.

conceptually b and c problems are different because (b) is 3 people are independently each other events then probability of 3 people is 0.00052 and

(c) is sample size =3 so that probability = 0.0077

e.

P(X1 > 200) = (200-165)/25

= 35/25 = 1.4

= P ( Z >1.4) From Standard Normal Table

= 0.0808

P(X2 > 200) = (200-165)/25

= 35/25 = 1.4

= P ( Z >1.4) From Standard Normal Table

= 0.0808

P(X3 > 200) = (200-165)/25

= 35/25 = 1.4

= P ( Z >1.4) From Standard Normal Table

= 0.0808

P(X4 > 200) = (200-165)/25

= 35/25 = 1.4

= P ( Z >1.4) From Standard Normal Table

= 0.0808

P(X5 > 200) = (200-165)/25

= 35/25 = 1.4

= P ( Z >1.4) From Standard Normal Table

= 0.0808

mutually exclusive events

p(X1+X2+X3+X4+X5)>1000 = p(X1>200)+p(X2>200)+p(X3>200)+p(X4>200)+p(X5>200)

=0.0808+0.0808+0.0808+0.0808+0.0808 = 0.404

p(not A) = 1-p(A)

p( not (X1+X2+X3+X4+X5)>1000) = 1-0.404 = 0.596

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