he citizens of Townopolis have normally distributed body weights with a mean of
ID: 2933801 • Letter: H
Question
he citizens of Townopolis have normally distributed body weights with a mean of 204 lbs. and a standard deviation of 22 lbs. If an elevator in town has a maximum capacity of 40 persons and a rated maximum weight of 8600 lbs., what is the probability that the elevator, when filled to capacity, will not exceed its rated weight? (in other words, what is the probability that a sample of 40 citizens will have a mean weight of less than 215 lbs.?) Give your answer to 4 decimal places.
Suppose that ice cream consumption per person at parties is normally distributed with a mean of 0.35 gallons, and a standard deviation of 0.11 gallons. If you are throwing a party with 55 guests, how much ice cream do you need to buy to make sure that the probability of having enough is 90%? Round up to the next gallon. Do not use units in your answer.
he mean salary of the 1200 employees of MegaCom is $58,400 with a standard deviation of $6,800. What is the probability that a random sample of 140 employees of MegaCom will have a salary of less than $58,200? Round your answer to 5 decimal places.
Explanation / Answer
Q1.
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 ) = 204
standard Deviation ( sd )= 22/ Sqrt ( 40 ) =3.4785
sample size (n) = 40
a.
P(X < 215) = (215-204)/22/ Sqrt ( 40 )
= 11/3.4785= 3.1623
= P ( Z <3.1623) From Standard NOrmal Table
= 0.9992
Q2.
mean ( u ) = 0.35
standard Deviation ( sd )= 0.11/ Sqrt ( 55 ) =0.0148
sample size (n) = 55
P ( Z > x ) = 0.1
Value of z to the cumulative probability of 0.1 from normal table is 1.2816
P( x-u / (s.d) > x - 0.35/0.0148) = 0.1
That is, ( x - 0.35/0.0148) = 1.2816
--> x = 1.2816 * 0.0148+0.35 = 0.369
Q3.
mean ( u ) = 58400
standard Deviation ( sd )= 6800/ Sqrt ( 140 ) =574.7049
sample size (n) = 140
P(X < 58200) = (58200-58400)/6800/ Sqrt ( 140 )
= -200/574.7049= -0.348
= P ( Z <-0.348) From Standard NOrmal Table
= 0.3639
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