[21 points] Suppose that 10-ft lengths of a certain type of cable have breaking
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Question
[21 points] Suppose that 10-ft lengths of a certain type of cable have breaking strengths that are normally distributed with mean 2. 450 lb and standard deviation = 50 lb (a) Find the probability that one such cable will have a strength greater than 536 lb. [4 points] Let X the mean breaking strength for a random sample of nine such cables. Clearly, the value of X will vary from one sample to another. Describe its sampling distribution by giving the (i) shape, (ii) mean, and (iii) standard deviation of the distribution. [1 point each] Find the probability that the sample mean of nine cables will be between 423 and 480. [4 points] (d Find the probability that the sample mean of 35 cables will be less than 428. [4 points] (e) Suppose that the distribution of breaking strengths for all cables in the population had been non-normal or unknown. [2 points each] (i) Could part (a) have been solved using the information given? Why or why not? (ii) Could part (c) have been solved using the information given? Why or why not? (iii) Could part (d) have been solved using the information given? Why or why not? [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 3 more than 200 lb? (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), ie. how are the events (e) Have you ever ridden in an elevator and read a sign stating its maximum load and Il three weigh more than 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, +...+X, > 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
2.e.
i)
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 ) = 450
standard Deviation ( sd )= 50/ Sqrt ( 1 ) =50
sample size (n) = 1
P(X > 536) = (536-450)/50/ Sqrt ( 1 )
= 86/50= 1.72
= P ( Z >1.72) From Standard Normal Table
= 0.0427
ii)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 423) = (423-450)/50/ Sqrt ( 9 )
= -27/16.6667
= -1.62
= P ( Z <-1.62) From Standard Normal Table
= 0.0526
P(X < 480) = (480-450)/50/ Sqrt ( 9 )
= 30/16.6667 = 1.8
= P ( Z <1.8) From Standard Normal Table
= 0.9641
P(423 < X < 480) = 0.9641-0.0526 = 0.9115
iii)P(X < 428) = (428-450)/50/ Sqrt ( 35 )
= -22/8.4515= -2.6031
= P ( Z <-2.6031) From Standard NOrmal Table
= 0.0046
distribution of breaking strenght for all cables in the population had been known each part of a,c,d
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