SOLUTION: Assume that the readings on the thermometers are normally distributed with a mean of 0 degree and a standard deviation of 1.00 degree. Find the indicated probability, where z is th

Algebra ->  Probability-and-statistics -> SOLUTION: Assume that the readings on the thermometers are normally distributed with a mean of 0 degree and a standard deviation of 1.00 degree. Find the indicated probability, where z is th      Log On


   



Question 1206577: Assume that the readings on the thermometers are normally distributed with a mean of 0 degree and a standard deviation of 1.00 degree. Find the indicated probability, where z is the reading in degrees.
a) Between 0 and 1.96
b) Less than -1.47
c) Between 0.89 and 1.78
d) Greater than -1.05

Answer by ikleyn(52855) About Me  (Show Source):
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Assume that the readings on the thermometers are normally distributed
with a mean of 0 degree and a standard deviation of 1.00 degree.
Find the indicated probability, where z is the reading in degrees.
a) Between 0 and 1.96
b) Less than -1.47
c) Between 0.89 and 1.78
d) Greater than -1.05
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

(a)  This probability is the area under the given/described normal curve 
     between the z-marks z1 = 0 and z2 = 1.96.

     Use a regular calculator TI-83/84 and its standard function normcdf.

                      z1    z2   mean   SD      <<<---=== formatting pattern
         P = normcdf( 0,   1.96,  0,   1.00).


     You will get the value  P = 0.475   (rounded).    ANSWER

     The same can be done using an online (free of charge) calculator
     https://onlinestatbook.com/2/calculators/normal_dist.html



(b)  This probability is the area under the given/described normal curve 
     on the left of the z-mark z2 = -1.47.

     Use a regular calculator TI-83/84 and its standard function normcdf.

                      z1        z2   mean   SD      <<<---=== formatting pattern
         P = normcdf( -9999,  -1.47,  0,   1.00).


     You will get the value  P = 0.0708   (rounded).    ANSWER

     The same can be done using an online (free of charge) calculator
     https://onlinestatbook.com/2/calculators/normal_dist.html



(c)  This case is similar (a TWIN) to case (a).  Do it as instructed in (a),
     with obvious modifications.



(d)  This probability is the area under the given/described normal curve 
     on the right of the z-mark z1 = -1.05.

     Use a regular calculator TI-83/84 and its standard function normcdf.

                        z1     z2   mean   SD      <<<---=== formatting pattern
         P = normcdf( -1.05,  9999,  0,   1.00).


     You will get the value  P = 0.8531   (rounded).    ANSWER

     The same can be done using an online (free of charge) calculator
     https://onlinestatbook.com/2/calculators/normal_dist.html

Solved.