Question

In: Statistics and Probability

The monthly utility bills in a certain city are normally distributed with a mean of $100...

The monthly utility bills in a certain city are normally distributed with a mean of $100 and standard deviation of $12.
a) A utility bill is randomly selected. Find the probability that it is: i) less than $80
ii) between $75 and $ 115
b) What percentage of the utility bills are for more than $125?
c) If 300 utility bills were selected at random, how many would be less than $90?
d) The utility company wants to give a small prize to those customers who do the most to conserve energy. They will give this only to the 15% of customers with the lowest utility bills. What is the highest utility bill one can have and still receive the prize?
e) Forty-one utility bills are selected at random. Determine the probability that the average bill amount is greater than $107

Solutions

Expert Solution

a) i)

µ =    100          
σ =    12          
              
P( X ≤    80   ) = P( (X-µ)/σ ≤ (80-100) /12)      
=P(Z ≤   -1.667   ) =   0.04779   (answer)
.

ii)

µ =    100                              
σ =    12                              
we need to calculate probability for ,                                  
P (   75   < X <   115   )                  
=P( (75-100)/12 < (X-µ)/σ < (115-100)/12 )                                  
                                  
P (    -2.083   < Z <    1.250   )                   
= P ( Z <    1.250   ) - P ( Z <   -2.083   ) =    0.8944   -    0.0186   =    0.8757
...........

b)

µ =    100                  
σ =    12                  
                      
P ( X ≥   125.00   ) = P( (X-µ)/σ ≥ (125-100) / 12)              
= P(Z ≥   2.083   ) = P( Z <   -2.083   ) =    0.0186 = 1.86% (answer)

.........

c)

µ =    100      
σ =    12      
          
P( X ≤    90   ) = P( (X-µ)/σ ≤ (90-100) /12)  
=P(Z ≤   -0.833   ) =   0.20233
the number of bills less than 90 will be = 300*0.20233

= 60.699 ~ 61 (round up)

d)

µ=   100                  
σ =    12                  
proportion=   0.15                  
                      
Z value at    0.15   =   -1.04   (excel formula =NORMSINV(   0.15   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   -1.04   *   12   +   100  
X   =   87.56   (answer)          

thanks

revert back for doubt


Related Solutions

in a certain city, the monthly water bill amount is normally distributed with mean 25 and...
in a certain city, the monthly water bill amount is normally distributed with mean 25 and standard deviation 15. A sample of 36 bills was selected. What is the probability that the average water bill amount for the sample was between 26.7 and 30? round your answer to 4 decimal places. Do not write as percentage.
The value of investment properties in a certain city is normally distributed with a mean of...
The value of investment properties in a certain city is normally distributed with a mean of $360,000 and standard deviation of $60,000. (a) What is the probability that a randomly selected property is worth less than $250,000? (b) If the value of investment properties for 5% of all investors is less than a given amount, what is the maximum amount that an investor would expect to pay for a property? (c) If a random sample of 45 investors is selected,...
The scores for a certain test of intelligence are normally distributed with mean 100 and standard...
The scores for a certain test of intelligence are normally distributed with mean 100 and standard deviation Find the 80th percentile of these scores. Below is the table used, I still cant figure out what the 80% would be here: Standard Scores and Percentiles ​z-score Percentile ​z-score Percentile ​z-score Percentile ​z-score Percentile minus−3.5 00.02 minus−1.00 15.87 0.00 50.00 1.1 86.43 minus−3.0 00.13 minus−0.95 17.11 0.05 51.99 1.2 88.49 minus−2.9 00.19 minus−0.90 18.41 0.10 53.98 1.3 90.32 minus−2.8 00.26 minus−0.85 19.77...
In a certain winter vacation city, snowfalls in January are Normally distributed with mean = 175.8...
In a certain winter vacation city, snowfalls in January are Normally distributed with mean = 175.8 inches and standard deviation = 25.8 inches. A researcher randomly selects 36 snowfalls. What is the probability that the mean of the sample is less than 170 inches? What is the probability that the mean of the sample is greater than 180 inches?
[40 pts] Consider a certain city with population 50000, where monthly income is normally distributed with...
[40 pts] Consider a certain city with population 50000, where monthly income is normally distributed with mean 2000TL and standard deviation 300TL. (you may use the table below for calculations) [20pts] Find the maximum income among the people with the lowest 10% income. [20pts] Find the minimum income among the people with the highest 30% income.
The prices of condos in a city are normally distributed with a mean of $100,000 and...
The prices of condos in a city are normally distributed with a mean of $100,000 and a standard deviation of $32,000. Answer the following questions rounding your solutions to 4 decimal places. 1. The city government exempts the cheapest 6% of the condos from city taxes. What is the maximum price of the condos that will be exempt from city taxes? 2. If 2% of the most expensive condos are subject to a luxury tax, what is the minimum price...
The weights of the fish in a certain lake are normally distributed with a mean of...
The weights of the fish in a certain lake are normally distributed with a mean of 20 lb and a standard deviation of 9. If 9 fish are randomly selected, what is the probability that the mean weight will be between 17.6 and 23.6 lb? Write your answer as a decimal rounded to 4 places.
The weights of the fish in a certain lake are normally distributed with a mean of...
The weights of the fish in a certain lake are normally distributed with a mean of 9.4 lb and a standard deviation of 2.3. If 42 fish are randomly selected, what is the probability that the mean weight will be more than 9.7 lb?
The weights of the fish in a certain lake are normally distributed with a mean of...
The weights of the fish in a certain lake are normally distributed with a mean of 13 pounds and a standard deviation of 6. If a sample of 9 fish are randomly selected, what is the probability that the mean weight will be between 10.2 and 16.6 pounds?
The weights of the fish in a certain lake are normally distributed with a mean of...
The weights of the fish in a certain lake are normally distributed with a mean of 9.9 lb and a standard deviation of 2.1. If 75 fish are randomly selected, what is the probability that the mean weight will be between 7.7 and 10.4 lb?
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT