Question

In: Statistics and Probability

23. NOTE: Answers using z-scores rounded to 3 (or more) decimal places will work for this...

23. NOTE: Answers using z-scores rounded to 3 (or more) decimal places will work for this problem.

The population of weights for men attending a local health club is normally distributed with a mean of 175-lbs and a standard deviation of 26-lbs. An elevator in the health club is limited to 32 occupants, but it will be overloaded if the total weight is in excess of 5952-lbs.

a. Assume that there are 32 men in the elevator. What is the average weight beyond which the elevator would be considered overloaded?
average weight = lbs

b. What is the probability that one randomly selected male health club member will exceed this weight?
P(one man exceeds) =  
(Report answer accurate to 4 decimal places.)

c. If we assume that 32 male occupants in the elevator are the result of a random selection, find the probability that the elevator will be overloaded?
P(elevator overloaded) =  
(Report answer accurate to 4 decimal places.)

d. If the elevator is full (on average) 6 times a day, how many times will the elevator be overloaded in one (non-leap) year?
number of times overloaded =  
(Report answer rounded to the nearest whole number.)

Solutions

Expert Solution

Solution:

The population of weights for men attending a local health club is normally distributed with a mean of 175-lbs and a standard deviation of 26-lbs. An elevator in the health club is limited to 32 occupants, but it will be overloaded if the total weight is in excess of 5952-lbs.

a.

Assume that there are 32 men in the elevator.

We have to find the average weight beyond which the elevator would be considered overloaded.

The elevator will be overloaded if the total weight is in excess of 5952-lbs.

So, the average weight would be =

Answer: The beyond the average weight 186-lbs elevator would be considered overloaded.

b.

Let X be the weight of a randomly chosen man from the given population.

We are given that,

Hence,

So, the probability that one randomly selected male health club member will exceed the weight 186-lbs is,

Answer: The probability that one randomly selected male health club member will exceed the weight 186-lbs is 0.3361.

c.

We assume that 32 male occupants in the elevator are the result of a random selection, we have to find the probability that the elevator will be overloaded.

We know the elevator would be considered overloaded if the average weight is more than 186-lbs.

The distribution of the sample mean is given by,

In our data,

So,

Hence,

The probability that the lift is overloaded is given by,

Answer: The probability that the lift is overloaded is 0.0083.

d. The elevator is full (on average) 6 times a day. The number of times the elevator will be overloaded in one (non-leap) year is = 6*36 = 216.


Related Solutions

NOTE: Answers using z-scores rounded to 3 (or more) decimal places will work for this problem....
NOTE: Answers using z-scores rounded to 3 (or more) decimal places will work for this problem. The population of weights for men attending a local health club is normally distributed with a mean of 179-lbs and a standard deviation of 29-lbs. An elevator in the health club is limited to 34 occupants, but it will be overloaded if the total weight is in excess of 6460-lbs. Assume that there are 34 men in the elevator. What is the average weight...
Answer the following break even problems. In addition to the answers (rounded to 0 decimal places),...
Answer the following break even problems. In addition to the answers (rounded to 0 decimal places), show the spreadsheets you created to solve them (like you do for homeworks in lab). Next to any cells with equations, put the equation in text. For example, if I was doing a spreadsheet to sum the number of fish my son and I caught, it would look like: Q1 (5 points) Stew’s Plastics produces a variety of CD cases. The best-selling product is...
Enter answers rounded to 5 decimal places. 1. In pharmacologic research a variety of clinical chemistry...
Enter answers rounded to 5 decimal places. 1. In pharmacologic research a variety of clinical chemistry measurements are routinely monitored closely for evidence of side effects of the medication under study. Suppose typical blood glucose levels are normally distributed with mean 90 mg/dL and standard deviation 38 mg/dL. 1a. If the normal range is 65-120 mg/dL, that what percentage of values will fall into the normal range? 1b. In some studies only values that are at least 1.5 times as...
Note: Each bound should be rounded to three decimal places. Q: A random sample of n=100...
Note: Each bound should be rounded to three decimal places. Q: A random sample of n=100 observations produced a mean of x⎯⎯⎯=35 with a standard deviation of s=5. (a) Find a 95% confidence interval for μ Lower-bound: Upper-bound: (b) Find a 90% confidence interval for μ Lower-bound: Upper-bound: (c) Find a 99% confidence interval for μ Lower-bound: Upper-bound:
Find the price (per $100 face value, rounded to 3 decimal places) of a 12% Treasury...
Find the price (per $100 face value, rounded to 3 decimal places) of a 12% Treasury bond, 145 days before maturity, at a yield of 6.26% p.a. 1
Assume that z is the test statistic. (Give your answers correct to two decimal places.) (a)...
Assume that z is the test statistic. (Give your answers correct to two decimal places.) (a) Calculate the value of z for Ho: μ = 10, σ = 2.8, n = 36, x = 11.4. _________ (b) Calculate the value of z for Ho: μ = 120, σ = 26, n = 26, x = 125.9. _________ (c) Calculate the value of z for Ho: μ = 18.2, σ = 4.4, n = 140, x = 18.88. __________ (d) Calculate...
Express numerical answers in decimal form and round to 3 decimal places as needed (unless otherwise...
Express numerical answers in decimal form and round to 3 decimal places as needed (unless otherwise stated). [8] 1) The homework scores (out of 10 points) for a sample of 9 students are listed: 1, 7, 7, 8, 9, 9, 10, 10, 10 a) Find the five number summary (whole numbers, in order). Five number summary: , , , ,
Use the z-table to find the requested probabilities. Enter your answers to 4 decimal places. (a)    ...
Use the z-table to find the requested probabilities. Enter your answers to 4 decimal places. (a)     P(z < −2.55) = (b)     P(z ≥ 2.04) = (c)     P(−1.92 < z < 1.93) =
Use the z-table to find the requested probabilities. Enter your answers to 4 decimal places. (a)    ...
Use the z-table to find the requested probabilities. Enter your answers to 4 decimal places. (a)     P(z < 1.26) = (b)     P(z ≥ −1.71) = (c)     P(−2.19 < z < 2.65) =
Determine the value z* that satisfies the conditions below. (Round all answers to two decimal places.)...
Determine the value z* that satisfies the conditions below. (Round all answers to two decimal places.) (a) Separates the largest 3.2% of all z values from the others z* =   (b) Separates the largest 0.8% of all z values from the others z* =   (c) Separates the smallest 5.6% of all z values from the others z* = (d) Separates the smallest 12.1% of all z values from the others z* =
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT