Question

In: Statistics and Probability

The average daily rate of a hotel in Canada as of August 2018 was $172.75. Assume...

The average daily rate of a hotel in Canada as of August 2018 was $172.75. Assume the average daily rate follows a normal distribution with a standard deviation of $28.60.

Standard Normal Distribution Table ( note you need access to this table in order to do this question)

a. What is the probability that the average daily rate of a Canadian hotel will be:

(i) less than $175
P(X < 175)=P(X < 175)=



(ii) more than $205
P(X > 205)=P(X > 205)=



(iii) Between $145 and $185
P(145 < X < 185)=P(145 < X < 185)=

b. Determine the average daily rates that separate the:

(i) top 7% of average daily rates from the rest of the daily rates or from the bottom 93% of average daily rates
x=x=

Round to 2 decimal places.


(ii) bottom 25% of average daily rates from the rest of the daily rates
x=x=

Round to 2 decimal places.


(iii) middle 70% of average daily rates from the rest of the daily rates

  < x <  < x <  

Round to 2 decimal places.

h(i,x)=

{10if  program i halts on input x,otherwise.{1if  program i halts on input x,0otherwise.

Solutions

Expert Solution

Solution :

Given that ,

mean = = 172.75

standard deviation = = 28.60

(i)  P(x < 175 ) = P[(x - ) / < ( 175 - 172.75 ) / 28.60 ]

= P(z < 0.08)

Using z table,

= 0.5319

Probability = 0.5319

(ii)

P(x > 205 ) = 1 - P( x < 205)

=1- P[(x - ) / < ( 205 - 172.75 ) / 28.60 ]

=1- P(z < 1.13 )

Using z table,

= 1 - 0.8708

= 0.1292

Probability = 0.1292

(iii)

P( 145 < x < 185)

= P[( 145 - 172.75 ) / 28.60 ) < (x - ) /  < ( 185 - 172.75 ) / 28.60 ) ]

= P( -0.97 < z < 0.43 )

= P(z < 0.43 ) - P(z < -0.97)

Using z table,

= 0.6664 - 0.1660

= 0.5004

Probability = 0.5004


(i)

The z - distribution of the 7% is

P(Z > z) = 7%

= P(Z < z ) = 0.07

= 1 - P(Z < z) = 1 - 0.07

= P(Z < z ) = 0.93

= P(Z < 1.476 ) = 0.93

z = 1.476

Using z-score formula,

x = z * +

x = 1.476 * 28.60 + 172.75

x = 214.9636

x = 214.96


(ii)

The z - distribution of the 25% is

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.6745 ) = 0.25

z = -0.6745

Using z-score formula,

x = z * +

x = -0.6745 * 28.60 + 172.75

x = 153.4593

x = 153.46


(iii)

Middle 70%

= 1 -70%  

= 1 - 0.70 = 0.30

/2 = 0.15

1 - /2 = 1 - 0.15 = 0.85

Z/2 = Z0.15 = -1.036

Z1 - /2 = Z0.85 = 1.036

Using z-score formula,

x = z * +

x = -1.036 * 28.60 + 172.75

x = 143.12

Using z-score formula,

x = z * +

x = 1.036 * 28.60 + 172.75

x = 202.38

Answer = 143.12 < x < 202 .38


Related Solutions

The average daily rate of a hotel in Canada as of August 2018 was $182.65. Assume...
The average daily rate of a hotel in Canada as of August 2018 was $182.65. Assume the average daily rate follows a normal distribution with a standard deviation of $25.70. Standard Normal Distribution Table a. What is the probability that the average daily rate of a Canadian hotel will be: (i) less than $175 P(X < 175)=P(X < 175)= (ii) more than $200 P(X > 200)=P(X > 200)= (iii) Between $150 and $190 P(150 < X < 190)=P(150 < X...
The average daily rate of a hotel in Canada as of August 2018 was $184.85. Assume...
The average daily rate of a hotel in Canada as of August 2018 was $184.85. Assume the average daily rate follows a normal distribution with a standard deviation of $27.70. Standard Normal Distribution Table a. What is the probability that the average daily rate of a Canadian hotel will be: (i) less than $165 P(X < 165)=P(X < 165)= (ii) more than $215 P(X > 215)=P(X > 215)= (iii) Between $160 and $190 P(160 < X < 190)=P(160 < X...
In exercise 12, the following data on x = average daily hotel room rate and y...
In exercise 12, the following data on x = average daily hotel room rate and y = amount spent on entertainment (The Wall Street Journal, August 18, 2011) lead to the estimated regression equation ŷ = 17.49 + 1.0334x. For these data SSE = 1541.4. Click on the webfile logo to reference the data. a. Predict the amount spent on entertainment for a particular city that has a daily room rate of $89 (to 2 decimals). $ b. Develop a...
Ada Hotel sells two room tpes: standard rooms and deluxe rooms. Average daily rate (ADR) and...
Ada Hotel sells two room tpes: standard rooms and deluxe rooms. Average daily rate (ADR) and variable costs (VC) of the two room types are provided in the table below: (Hint: Treat two room types as two different products.)                                                 ADR ($)   Variable Cost ($)               Standard rooms    394.50   256.43               Deluxe rooms   631.20   366.10                                     ...
Exercise 1: (10 marks) Average daily hotel room rate (x) and amount spent on entertainment (y)...
Exercise 1: Average daily hotel room rate (x) and amount spent on entertainment (y) is given in the following table for different cities. a.       Use these data to develop an estimated regression equation that could be used to predict the amount spent on entertainment for a given average daily hotel room rate. b.       What is the value of sample correlation coefficient? c.       Compute the coefficient of determination. What percentage of the variation in total amount spent on entertinment can be...
Problem 2. Ada Hotel sells two room tpes: standard rooms and deluxe rooms.  Average daily rate (ADR)...
Problem 2. Ada Hotel sells two room tpes: standard rooms and deluxe rooms.  Average daily rate (ADR) and variable costs (VC) of the two room types are provided in the table below: (Hint: Treat two room types as two different products.) ADR ($) Variable Cost ($) Standard rooms 461.20 299.78 Deluxe rooms 737.92 427.99 The Mock Hotel's fixed costs for a month is =          = 295168 Sales mix (contribution of each room type to total room revenue) of the hotel is: Deluxe...
A hotel chain wants to estimate the average number of rooms rented daily in each month....
A hotel chain wants to estimate the average number of rooms rented daily in each month. The population of rooms rented daily is assumed to be normally distributed for each month with a standard deviation of 24 rooms. During January, a sample of 16 days has a sample mean of 48 rooms. This information is used to calculate an interval estimate for the population mean to be from 40 to 56 rooms. What is the level of confidence of this...
A researcher wished to compare the average daily hotel room rates between San Francisco and Los...
A researcher wished to compare the average daily hotel room rates between San Francisco and Los Angeles. The researcher obtained an SRS of 15 hotels in downtown San Francisco and found the sample mean ? ̅1=$156, with a standard deviation ?_1= $15. The researcher also obtained an independent SRS of 10 hotels in downtown Los Angeles and found the sample mean ? ̅_2= $143, with a standard deviation ?_2= $10. Let 1 and 2 represent the mean cost of the...
(1) On August 1, 2018, We R Clean Company signed a 9-month contract with a hotel...
(1) On August 1, 2018, We R Clean Company signed a 9-month contract with a hotel chain to provide pool and spa cleaning services for 3 hotel sites. The contract price of $14,850 was collected on the date the contract was signed. The services will be provided evenly over the next 9 months, starting on August 1. The adjusting entry on December 31, 2018 will Credit Service Revenue for $6,600 Debit Earned Revenue for $6,600 Credit Service Revenue for 8,910...
Statistics Canada did a survey of Canadians in 2018 and determined that the average adult female...
Statistics Canada did a survey of Canadians in 2018 and determined that the average adult female over the age of 18 watches 25.6 hours of television per week while the average male watches 20.9 hours per week. Suppose that the sample comprised 35 females and 50 males (obviously the real samples were much larger) and that the sample standard deviations were 7.2 hours for females and 7 hours for males. a) [4 marks] Is there sufficient evidence, at the 5%...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT