1 answer

13-15 273 4 The Canal Limit Theorem Neto carry a load that is 255 greater than...

Question:

13-15
273 4 The Canal Limit Theorem Neto carry a load that is 255 greater than the state limit. The maimum capacity 5000 lb after i
273 4 The Canal Limit Theorem Neto carry a load that is 255 greater than the state limit. The maimum capacity 5000 lb after it is increased by 255. 27 male can have of up to 185 lb. If the elevator is looded with 27 adult male pa n id the hat it is overloaded because they have a mean weight greater than 15 th (As in w ethal weights of males are normally distributed with a means of a banda Bu a tion of 39 lb) Does this elevator appear to be safe? stor Safety Exercise we = 189 lb, which is bawdon Data Set Body Dua Repeat Exercise using = 174 Tb (instead of 189 Thi which is the assumed A l that was commonly used just a few years ago. What do you conclude about the using an outdmcan this w antially than it wil he Membership in Mensa requires a score in the top on a standard intelligence est der IQ test is designed for a mean of 100 and a standard deviation of 15, and scores ne Wed andla 10 DOTO The Elevator Safety totag Mensa Mersin ally distributed sed the minimum Wechsler IQ test score that satisfies the Mens requirement andomly selected adults take the Wechsler IQ test, find the probably that their mean costa 17 No when conceng Valmennt COS Table:903 These capacity of 20 pas hects take the Wechsler test and they have a mean of the individual cores are can we conclude that all 4 of them are eligible for Mensa? 12. Designing Manholes According to the website www.torchmale commanhole covers be a minimum of 22 in. in diameter, but can be as much as in. in diameter. As a manhole is constructed to have a circular opening with a diameter of 22 in. Men have der breadths that are normally distributed with a mean of 18.2 in and a standard deviation of 10 in (based on data from the National Health and Nutrition Examination Survey). What percentage of men will fit into the manhole! Assume that the Connecticut's Evensource company employs 36 men who work in man- boles. If 36 men are randomly selected, what is the probability that their mean shoulder breath is less than 18 5 in? Does this revalt suggest that money can be saved by making smaller man holes with a diameter of 18.5 in.? Why or why not? 11. Water Taxi Safety Passengers died when a water taxi sank in Baltimore's Inner Harbor Men are typically heavier than women and children, so when loading a water taxi, asuma worst case scenario in which all passengers are men. Assume that weights of men are normally distributed with a mean of 189 lb and a standard deviation of 39 b (based on Dua Set 1 "Body Data" in Appendix B). The water taxi that sank had a stated capacity of 25 passengers, and the boat was rated for a load limit of 3500 lb a. Given that the water taxi that sank was voted for a load limit of 3500 lb, what is the maxi mum mean weight of the passengers if the boat is filled to the stated capacity of 25 passengers! b. If the water taxi is filled with 25 randomly selected men, what is the probability that their me weight exceeds the value from part() c. After the water taxi sank, the weight assumptions were revised so that the new capacity be- came 20 passengers. If the water taxi is filled with 20 randomly selected men, what is the prob- ability that their mean weight exceeds 175 lb, which is the maximum mean weight that does because the total load to exceed 3500 in? d. Is the new capacity of 20 passengers safe? 14. Vending Machines Carters are now manufactured so that they have a mean weight of 5.670 g and a standard deviation of 0.062 and their weights are normally distributed. A vend ng machine is configured to accept only those quarters that weigh between 5 550 g and 5.790 a. If I randomly selected quarter is inserted into the vending machine, what is the probability that it will be accepted? continued Grading is too high 14 3 0471 Table Part because eduarters could not The curred one t ime, not in groups of 4

Answers

(13)

(a) Mean = \frac{\sum x}{n}

We need to the total weights of the passengers to be \leq3500 lb (less than or equal to 3500 lb)b. Since there are 25 passengers the mean will

\frac{\sum x}{n}= \frac{3500}{25} = 140 1Ь

(b) Let X\rightarrow weights of the men

X, Νίμ = 189, σ = 39)

According to the central limit, if X_{i} are independent and identically distributed variables, the mean distribution of these variables will follow normal distribution with parameters \mu and \frac{\sigma}{\sqrt{n}}

X25 • Nụ = 189, s/s */

39 X35 ~ N(189,

2 – score = I - 189 39/25

We need to find the probability that the mean weight is more than 140 lb.

P(\bar{X}>140)=P(Z>\frac{140-189}{39/\sqrt{25}})

  = P(Z > -6.282)

=  P(Z < 6.282)

= \mathbf{1} ...........Using normal dist tables, excel func 'normsdist(6.28)'

We can get 0.99999 but I am approximating to 1.

(c) Now instead of n= 25 we have n = 20. So we have the following distribution

\bar{X_{20}}\sim N(\mu=189,\frac{\sigma}{\sqrt{n}}=\frac{39}{\sqrt{20}})

\bar{X_{20}}\sim N(189,\frac{39}{\sqrt{20}})

P(\bar{X}>175)=P(Z>\frac{175-189}{39/\sqrt{20}})

=P(Z>-1.605)

=P(Z<1.605)

= \mathbf{0.9458}

Maximum mean weight

Total weight should not exceed 3500 lb

Max mean weight of 20 men = \frac{3500}{20}

= \mathbf{175}lb

(d) Since the probability that maximum mean weight 175 will be exceeded is 0.9458, which is too high, thats means the probability that total load will also have p=0.9458 of exceeding 3500 lb.

Therefore 20 passengers arrangements is not suitable since probability of overloading is high.

(14) Let X\rightarrowWeight of coins

X\sim N(\mu=5.670,\sigma=0.062)

z-score=\frac{x-5.670}{0.062}

The coin will be accepted only if it weighs between 5.550g - 5.790

Probability of coin being accepted

P(5.550<X<5.790)=P(X<5.790)-P(X<5.550)

   = P(Z < 5.790 – 5.670 0.069 ) - P(Z < 5,550 - 5.670 0.062

=P(Z<1.94)-P(Z<-1.94)

  =P(Z<1.94)-[1-P(Z<1.94)]

=97354-(1-0.97354)

= \mathbf{0.9471}

.

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