Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

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

ID: 3153968 • Letter: A

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

45.

z1 = lower z score =    -1.96      
z2 = upper z score =     1.96      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.024997895      
P(z < z2) =    0.975002105      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.95000421   [ANSWER]

********************

46.

Using a table/technology, the left tailed area of this is          
          
P(z <   1.645   ) =    0.950015094 [ANSWER]

*******************

47.

z1 = lower z score =    -2.575      
z2 = upper z score =     2.575      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.005012004      
P(z < z2) =    0.994987996      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.989975991   [ANSWER]

*******************

48.

z1 = lower z score =    -1.96      
z2 = upper z score =     1.96      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.024997895      
P(z < z2) =    0.975002105      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.95000421      

Thus, those outside this interval is the complement =    0.04999579   [ANSWER]  

*******************************************

Hi! Please submit the next part as a separate question. That way we can continue helping you! Please indicate which parts are not yet solved when you submit. Thanks!