Question

Find the following probabilities:

a) Pr{Z < 0.33}

b) Pr{Z ≥ −0.33}

c) Pr{−2.06 < Z < 2.06}

d) Pr{−2.06 < Z < 0.0}

e) Pr{−4.00 < Z < 0.0}

f) Pr{Z < −1.75 or Z > 1.75} (you want the probability that Z
is outside the range −1.75 to 1.75) g) Pr{−1.75 < Z <
1.75}

Answer #1

Solution:

Since you have posted a question with multiple sub-parts, we will solve first four sub-parts for you.

a) We have to find P( z < 0.33 ) = ............?

Using excel , =NORMSDIST(0.33)

P( z < 0.33 ) = 0.6293

b) We have to find P( Z > -0.33 ) = ............ ?

We know P( Z > z ) = 1- P( Z < z )

So, P( Z > -0.33 ) = 1- P( Z < -0.33 )

Using excel , =1-NORMSDIST(-0.33)

P( Z > -0.33 ) = 0.6293

c) P( -2.06 < Z < 2.06 ) = ..........?

Above probability can be written as,

P( -2.06 < Z < 2.06 ) = P( Z < 2.06 ) - P( Z < -2.06)

Using excel , =NORMSDIST(2.06)-NORMSDIST(-2.06)

P( -2.06 < Z < 2.06 ) = 0.9606

d) P( -2.06 < Z < 0.0 ) = ..........?

Above probability can be written as,

P( -2.06 < Z < 0.0 ) = P( Z < 0 ) - P( Z < -2.06)

Using excel , =NORMSDIST(0)-NORMSDIST(-2.06)

P( -2.06 < Z < 0.0 ) = 0.4803

Done

Find the following probabilities: Please show
work
a) Pr{Z < 0.33}
b) Pr{Z ≥ -0.33}
c) Pr{-1.67 < Z < 1.67}
d) Pr{-2.91 < Z < 0.0}
e) Pr{Z < -1.03 or Z > 1.03}
(you want the probability that Z is outside the range -1.03 to
1.03)

1) Find the following probabilities:
a) Pr{Z < 0.63}
b)Pr{Z ≥ −0.63}
c) Pr{−2.12 < Z < 2.12}
d) Pr{−2.12 < Z < 0.0}
e) Pr{−4.00 < Z < 0.0}
f) Pr{Z < −1.32 or Z > 1.32} (you want the probability that Z
is outside the range −1.32 to 1.32)
g) Pr{−1.32 < Z < 1.32} h) Add (f) and (g). Are you
surprised? Why or why not?

a) Pr{Z < 0.63}
b) Pr{Z ≥ −0.63}
c) Pr{−2.12 < Z < 2.12}
d) Pr{−2.12 < Z < 0.0}
e) Pr{−4.00 < Z < 0.0}
f) Pr{Z < −1.32 or Z > 1.32} (you want the probability
that Z is outside the range −1.32 to 1.32)
g) Pr{−1.32 < Z < 1.32}
h) Add (f) and (g). Are you surprised? Why or why not?

1) Find the following probabilities: a) Pr{Z < 0.67} b) Pr{Z
≥ -0.67} c) Pr{-2.05 < Z < 2.05} d) Pr{-2.91 < Z <
0.31} e) Pr{Z < -2.03 or Z > 2.03} (you want the probability
that Z is outside the range -3.03 to 3.03)
2) Assuming that for the height of women, μ = 65.2 inches and σ
= 2.9 inches, find the following: a) Pr{Y > 65.7} b) Pr{Y <
57.8} c) Pr{60 < Y < 69}...

Let X,Y,Z⊆U. If Pr(X)=0.21, Pr(Y)=0.33, Pr(Z)=0.39,
Pr(X∩Y)=0.09, Pr(X∩Z)=0.08, Pr(Y∩Z)=0.17, and Pr(X∩Y∩Z)=0.04, find
the following values:
Pr(X′∩Y∩Z′).
Pr(X′∩(Y∪Z′)).
Pr(X′).
Pr(X′∪Y∪Z′).

Find the following probabilities:
a. P (z > 1.96)
b. P (z > .96)
c. P (z > 3.00)
d. P (z < 1.96)
e. P (z < .49)

Find the following probabilities for the standard normal random
variable z. (Round your answers to four decimal places.)
(a) P(−1.43 < z < 0.64) =
(b) P(0.52 < z < 1.75) =
(c) P(−1.56 < z < −0.48) =
(d) P(z > 1.39) =
(e) P(z < −4.34) =

Given that z is a standard normal random variable, compute the
following probabilities. Round your answers to 4 decimal
places.
a. P(0 _< z _< 0.51)
b. P( -1.61 _< z _< 0)
c. P( z > 0.30)
d. P( z _> -0.31)
e. P( z < 2.06)
f. P( z _< -0.61)

Calculate the following probabilities using the standard normal
distribution. (Round your answers to four decimal places.)
(a)
P(0.0 ≤ Z ≤ 1.6)
(b)
P(−0.1 ≤ Z ≤ 0.0)
(c)
P(0.0 ≤ Z ≤ 1.49)
(d)
P(0.6 ≤ Z ≤ 1.51)
(e)
P(−2.05 ≤ Z ≤ −1.76)
(f)
P(−0.02 ≤ Z ≤ 3.54)
(g)
P(Z ≥ 2.60)
(h)
P(Z ≤ 1.66)
(i)
P(Z ≥ 6)
(j)
P(Z ≥ −8)

Find the following probabilities. (Round your answers to four
decimal places.)
(a) p(0 < z < 1.62)
(b) p(1.40 < z < 1.83)
(c) p(−0.38 < z < 1.55)
(d) p(z < −1.91)
(e) p(−1.32 < z < −0.86)
(f) p(z < 1.27)

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