Question #53370
Given z is a standard normal random variable, answer questions 1 through 4.
1. p(z = 2.3) is
a. .1056.
b. 0.
c. .8944.
d. .3944.
2. p(z ≥ -1.84) is
a. .5474.
b. 0.
c. .0329.
d. .9671.
3. p(z ≤ 1.4) is
a. .0808.
b. .9192.
c. .9927.
d. 0.
4. p(.5 ≤ z ≤ 2.9) is
a. 0.
b. .6915.
c. .3066.
d. 9981.
1. p(z = 2.3) is
a. .1056.
b. 0.
c. .8944.
d. .3944.
2. p(z ≥ -1.84) is
a. .5474.
b. 0.
c. .0329.
d. .9671.
3. p(z ≤ 1.4) is
a. .0808.
b. .9192.
c. .9927.
d. 0.
4. p(.5 ≤ z ≤ 2.9) is
a. 0.
b. .6915.
c. .3066.
d. 9981.
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#339914 on Dec 2023
#339914 on Dec 2023
Comments
The solution uses the standard normal distribution table in calculations (see https://math.arizona.edu/~rsims/ma464/standardnormaltable.pdf). There exist normal distribution calculators (for example, https://www.mathportal.org/calculators/statistics-calculator/normal-distribution-calculator.php). In this problem the standard normal random variable is given, hence the mean is 0 and the standard deviation is 1.
how do you punch the answer into the calculator?