Assume that thermometer readings are normally distributed with a mean of 0degree

Question

Assume that thermometer readings are normally distributed with a mean of 0degreeC and a standard deviation of 1.00degreeC. A thermometer is randomly selected and tested. In each case, draw a sketch, and find the temperature reading corresponding to the given information. 49. Find P_95, the 95th percentile. This is the temperature reading separating the bottom 95% from the top 5%. 50. Find P_1, the 1st percentile. This is the temperature reading separating the bottom 1% from the top 99%. 51. If 2.5% of the thermometers are rejected because they have readings that are too high and another 2.5% are rejected because they have readings that are too low, find the two readings that are cutoff values separating the rejected thermometers from the others. 52. If 0.5% of the thermometers are rejected because they have readings that are two low and another 0.5% are rejected because they have readings that are too high, find the two readings that are cutoff values separating the rejected thermometers from the others.

Explanation / Answer

49.

We get the z score from the given left tailed area. As      
      
Left tailed area =    0.95  
      
Then, using table or technology,      
      
z =    1.644853627   [ANSWER]

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50.

We get the z score from the given left tailed area. As      
      
Left tailed area =    0.01  
      
Then, using table or technology,      
      
z =    -2.326347874   [ANSWER]

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51.

FOR THE LOWER BOUND:

We get the z score from the given left tailed area. As      
      
Left tailed area =    0.025  
      
Then, using table or technology,      
      
z =    -1.959963985   [ANSWER]

FOR THE UPPER BOUND:

We get the z score from the given left tailed area. As      
      
Left tailed area = 1 - 0.025 =   0.975  
      
Then, using table or technology,      
      
z =    1.959963985   [ANSWER]

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52.

FOR THE LOWER BOUND:

We get the z score from the given left tailed area. As      
      
Left tailed area =    0.005  
      
Then, using table or technology,      
      
z =    -2.575829304   [ANSWER]

FOR THE UPPER BOUND:

We get the z score from the given left tailed area. As      
      
Left tailed area = 1 - 0.005 =   0.995  
      
Then, using table or technology,      
      
z =    2.575829304   [ANSWER]

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