Kaycee Roche

2021-02-08

Determine the area under the standard normal curve that lies between
(a) Upper Z equals -2.03 and Upper Z equals 2.03​,
​(b) Upper Z equals -1.56 and Upper Z equals 0​, and
​(c) Upper Z equals -1.51 and Upper Z equals 0.68.

AGRFTr

(a) We have:
$-2.03
Using the appendix's normal probability table, calculate the corresponding probability.
$P\left(Z<-2.03\right)$ is given in the row starting with -2.0 and in the column starting with .03 of the standard normal probability table in the appendix.
$P\left(Z<2.03\right)$ is given in the row starting with 2.0 and in the column starting with .03 of the standard normal probability table in the appendix.
The probability between two boundaries is then the difference between the probabilities to the left of the boundaries.
$P\left(-2.03
$=0.9788-0.0212$
$=0.9576$
Hence, the area under the normal distribution between  is approximately 0.9576.
(b) We have:
$-1.56
Using the appendix's normal probability table, calculate the corresponding probability.
$P\left(Z<-1.56\right)$ is given in the row starting with -1.5 and in the column starting with .06 of the standard normal probability table in the appendix.
$P\left(Z<0\right)$ is given in the row starting with 0.0 and in the column starting with .00 of the standard normal probability table in the appendix.
The probability between two boundaries is then the difference between the probabilities to the left of the boundaries.
$P\left(-1.56
$=0.5000-0.0594$
$=0.4406$
Hence, the area under the normal distribution between -1.56 and 0 is approximately 0.4406.
(c) We have:
$-1.51
Using the appendix's normal probability table, calculate the corresponding probability.
$P\left(Z<-1.51\right)$ is given in the row starting with -1.5 and in the column starting with .01 of the standard normal probability table in the appendix.
$P\left(Z<0.68\right)$ is given in the row starting with 0.6 and in the column starting with .08 of the standard normal probability table in the appendix.
The probability between two boundaries is then the difference between the probabilities to the left of the boundaries.
$P\left(-1.51
$=0.7517-0.0655$
$=0.6862$
Hence, the area under the normal distribution between -1.51 and 0.68 is approximately 0.6862.

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