Business Statistic Class Overall the room temperature for gu

Business Statistic Class

Overall the room temperature for guests at Beach Hotels chains is normally distributed around 73 degrees with a standard deviation of 2.0 degrees.

What’s the probability that the mean temperature will be between 68 and 72 degrees?

In looking to cut costs, the hotel decides to ban guests whose rooms are in the top 3%ile of temperatures. Which guests from problem E would no longer be allowed to stay?

As energysaving incentive, users at or below the

15%ile of temperatures will receive a free

timeshare. What temperature do these guests need to stay below?

How likely (what is the probability) is to set the

room temperature between 71 and 74?

How likely (what is the probability) is it that Frosty

the Snowman (name not previously mentioned)

will set his room temperature between 36 and 40?

What percentile does 73 room temperature rank at?

If 50 people (50 = size of the sample) stay at the

hotel, what’s the likelihood that their mean temperature will be within a degree of the population mean?

Solution

a)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    68      
x2 = upper bound =    72      
u = mean =    73      
          
s = standard deviation =    2      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2.5      
z2 = upper z score = (x2 - u) / s =    -0.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.006209665      
P(z < z2) =    0.308537539      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.302327873   [ANSWER]

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b)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.97      
          
Then, using table or technology,          
          
z =    1.880793608      
          
As x = u + z * s,          
          
where          
          
u = mean =    73      
z = the critical z score =    1.880793608      
s = standard deviation =    2      
          
Then          
          
x = critical value =    76.76158722   [ANSWER]

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c)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.15      
          
Then, using table or technology,          
          
z =    -1.036433389      
          
As x = u + z * s,          
          
where          
          
u = mean =    73      
z = the critical z score =    -1.036433389      
s = standard deviation =    2      
          
Then          
          
x = critical value =    70.92713322   [ANSWER]

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d)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    71      
x2 = upper bound =    74      
u = mean =    73      
          
s = standard deviation =    2      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1      
z2 = upper z score = (x2 - u) / s =    0.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.158655254      
P(z < z2) =    0.691462461      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.532807207   [ANSWER]  
  

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