2. Given that z is a standard normal random variable, compute the following probabilities.
P(-1 ≤ z ≤ 0) (Round to four decimal places) Answer
P(-1.5 ≤ z ≤ 0) (Round to four decimal places) Answer
P(-2 < z < 0) (Round to four decimal places) Answer
P(-2.5 < z < 0) (Round to four decimal places) Answer
P(-3 ≤ z ≤ 0) (Round to four decimal places)
3. Given that z is a standard normal random variable, compute the following probabilities.
P(0 ≤ z ≤ .83) (Round to four decimal places) Answer
P(-1.57 ≤ z ≤ 0) (Round to four decimal places) Answer
P(z > .44) (Round to four decimal places) Answer
P(z ≥ -.23) (Round to four decimal places) Answer
P(z < 1.20) (Round to four decimal places) Answer
P(z < -.71) (Round to four decimal places)
4.Given that z is a standard normal random variable, compute the following probabilities.
P(-1.98 ≤ z ≤ .49) (Round to four decimal places) Answer
P(.52 ≤ z ≤ 1.22) (Round to four decimal places) Answer
P(-1.75 ≤ z ≤ -1.04) (Round to four decimal places)
5. Given that z is a standard normal random variable, find z for each situation.
The area to the right of z is .1314. (Round to two decimal places) Answer
The area to the left of z is .6700. (Round to two decimal places) Answer
The area to the right of z is .01. (Round to two decimal places) Answer
The area to the right of z is .025. (Round to two decimal places)
from normal value table:
2)
a) P(-1 ≤ z ≤ 0) =0.5000-0.1587=0.3413
b) P(-1.5 ≤ z ≤ 0) =0.5-0.0668=0.4332
c) P(-2 < z < 0) =0.5-0.0228=0.4772
d) P(-2.5 < z < 0) =0.5-0.0062=0.4938
e) P(-3 ≤ z ≤ 0)=0-0.0013=0.4987
3) a)P(-1.57 ≤ z ≤ 0) =0.5-0.0582=0.4418
b) P(0 ≤ z ≤ .83) =0.7967-0.5=0.2967
c) P(z > .44) =0.3300
d) P(z ≥ -.23) =0.5910
e) P(z < 1.20) =0.8849
f) P(z < -.71) =0.2389
4)
a) P(-1.98 ≤ z ≤ .49) =0.6879-0.0239=0.6640
b) P(.52 ≤ z ≤ 1.22) =0.8888-0.6985=0.1903
c) P(-1.75 ≤ z ≤ -1.04) =0.1492-0.0401=0.1091
5) The area to the right of z is .1314 ; z score=1.12
The area to the left of z is .6700 ; z score=0.44
The area to the right of z is .01 ; z score=2.33
The area to the right of z is .025 ; z score=1.96
2. Given that z is a standard normal random variable, computethe following probabilities. P(-1 ≤...
13. Given that z is a standard normal random variable, compute the following probabilities a. P(-1.98 z .49) b. P(.52szs 1.22) c. P(-1.75-z<-1.04)
4) Given that z is a standard normal random variable, find z for each situation The area to the left of z is .9750. The area between 0 and z is .4750. The area to the left of z is .7291. The area to the right of z is .1314. The area to the left of z is .6700. The area to the right of z is .3300.
4) Given that z is a standard normal random variable, find z for each situation (using excel): The area to the left of z is .9750. The area between 0 and z is .4750. The area to the left of z is .7291. The area to the right of z is .1314. The area to the left of z is .6700. The area to the right of z is .3300.
1. Given that z represents a standard normal random variable, provide the probabilities for the following: P (z < 1.56) _____________ P (z > -.94) _____________ P (1.57 < z < 2.30) ____________
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