Assume that thermometer readings are normally distributed with a mean of 0°c and standard deviation of 1.00°c. a thermometer is randomly selected and tested. for the case below, draw a sketch, and find the probability of the reading. (the given values are in celsius degrees)

between 0.50 and 2.00

the probability of getting a reading between 0.50c and 2.00c is

  0.2857

Step-by-step explanation:

The probability of interest is the area under the probability density curve between the z-values associated with the temperature limits of interest.

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The first attachment shows a "sketch" of the distribution and the area of the portion of interest. (It also shows the probability as 0.2858.)

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The second attachment shows the table values of interest for this problem. The z-values that we want to look for in the table are ...

  z1 = (0.50° -0°)/1.00° = 0.50

and

  z2 = (2.00° -0°)/1.00° = 2.00

The area of the probability density function to the left of each of these z-values is given in the table, so the area between them is the difference of table values:

  0.9772 -0.6915 = 0.2857

The probability of a reading between 0.50 and 2.00 is about 0.2857.

5y

step-by-step explanation: