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Given that z is a standard normal random variable, find z for each situation (to 2 decimals)a. The area to the left of z is 0.209. (Enter negative value as negative number.)b. The area between -z and z is 0.903-z and z is 0.2206c. The area betweend. The area to the left of z is 0.9953 .e. The area to the right of z is 0.6915. (Enter negative value as negative number.)

Question
Given that z is a standard normal random variable, find z for each situation (to 2 decimals)
a. The area to the left of z is 0.209. (Enter negative value as negative number.)
b. The area between -z and z is 0.903
-z and z is 0.2206
c. The area between
d. The area to the left of z is 0.9953 .
e. The area to the right of z is 0.6915. (Enter negative value as negative number.)
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Given that z is a standard normal random variable, find z for each situation (to 2 decimals) a. The area to the left of z is 0.209. (Enter negative value as negative number.) b. The area between -z and z is 0.903 -z and z is 0.2206 c. The area between d. The area to the left of z is 0.9953 . e. The area to the right of z is 0.6915. (Enter negative value as negative number.)

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Step 1

a.

The area to the left of z is 0.209 can be written as P(Z < z)=0.209.

In the z-table for negative values of z, the z-value corresponding to the nearest probability of 0.209 is -0.81.

Answer: -0.81

 

b.

The area between –z and z is 0.903 can be written as P(-z <Z <z) = 0.903.

P(-Z<)P(Z <-)-P(Z<--)
P(Z<)-1-P(Z<:)]
-2P (Z <)-
2P(Z<)-1 0.903
1+0.903
P(Z<E)
2
P(Z<) 0.9515
In the -table, the z-value corresponding to the nearest probability of
0.9515 is 1.66
=-1.66| and |- = 1.66
Answer:
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P(-Z<)P(Z <-)-P(Z<--) P(Z<)-1-P(Z<:)] -2P (Z <)- 2P(Z<)-1 0.903 1+0.903 P(Z<E) 2 P(Z<) 0.9515 In the -table, the z-value corresponding to the nearest probability of 0.9515 is 1.66 =-1.66| and |- = 1.66 Answer:

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Step 2

c.

The area between –z and z is 0.2206 can be written as P(-...

P(-Z <)2P(Z <)-1
2P(Z <)1 0.2206
1+0.2206
P(Z<E)
2
P(Z <0.6103
In the -table, the z-value corresponding to the nearest probability of
0.6103 is 0.28.
=-0.28| and |: = 0.28
Answer:
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P(-Z <)2P(Z <)-1 2P(Z <)1 0.2206 1+0.2206 P(Z<E) 2 P(Z <0.6103 In the -table, the z-value corresponding to the nearest probability of 0.6103 is 0.28. =-0.28| and |: = 0.28 Answer:

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