.
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.