Question

Let *Z* be a standard normal random variable (mean = 0
and sd = 1) and calculate the following probabilities:

(a) *Pr*(0 ≤ *Z* ≤
2.68)

(b) *Pr*(0 ≤ *Z* ≤ 2)

(c) * Pr*(−2.60 ≤

(d) * Pr*(−2.60 ≤

(e) *Pr*(*Z* ≤ 1.26)

Answer #1

Let Z be a standard normal random variable (mean = 0
and sd = 1) and calculate the following probabilities:
(a) Pr(0 ≤ Z ≤
2.49)
(b) Pr(0 ≤ Z ≤ 1)
(c)
Pr(−2.50 ≤ Z ≤ 0)
(d)
Pr(−2.50 ≤ Z ≤ 2.50)
(e) Pr(Z ≤ 1.52)

Let z be a normal random variable with mean 0 and
standard deviation 1. What is
P(-2.25 < z < -1.1)?
a 0.3643
b 0.8643
c 0.1235
d 0.4878
e 0.5000
Let zbe a normal random variable with mean 0 and
standard deviation 1. The 50thpercentile of zis
____________.
a 0.6700
b -1.254
c 0.0000
d 1.2800
e 0.5000
Let zbe a normal random variable with mean 0 and
standard deviation 1. The 75thpercentile of zis
____________.
a 0.6700
b...

Let z denote a random variable having a normal
distribution with μ = 0 and σ = 1. Determine each
of the probabilities below. (Round all answers to four decimal
places.)
(a) P(z < 0.1) =
(b) P(z < -0.1) =
(c) P(0.40 < z < 0.85) =
(d) P(-0.85 < z < -0.40)
=
(e) P(-0.40 < z < 0.85) =
(f) P(z > -1.26) =
(g) P(z < -1.49 or z > 2.50) =

Find the following probabilities for the standard normal random
variable ?:
(a) ?(−0.81≤?≤0.08)=
(b) ?(−0.57≤?≤1.05)=
(c) ?(?≤1.35)=
(d) ?(?>−1.26)=
HINT: The standard normal random variable has a mean of 0 and
standard deviation of 1 and is represented by z.

-. For what follows, Z denotes the standard normal random
variable.
(1) P(Z<1.26) is about
(a) 0.7924
(b) 0.3962
(c) 0.1038
(d) 0.6038
(e) 0.8962
(2) P(-0.54 < Z < 0.78) is about
(a) 0.2054
(b) 0.0769
(c) 0.4877
(d) 0.2823
(e) 0.6877
-Assume the random variable X is normally distributed with mean
172 and standard deviation 10. Find the following
probabilities.
(3) P(X<160)
(a) 0.8849
(b) 0.1151
(c) 0.9999
(d) 0.8078
(e) 0.8438
(4) P(X>150)
(a) 0.0139
(b) 0.9861...

Let Z ∼ N(0, 1) be a standard normal random variable, and let z
be a possible
value of Z. How large would z have to be to be considered an
outlier by the 1.5 IQR rule?

Let Z be a standard normal random variable and calculate the
following probabilities, drawing pictures wherever appropriate.
(Round your answers to four decimal places.)
(a) P(0 ≤ Z ≤ 2.57)
(b) P(0 ≤ Z ≤ 2)
(c) P(−2.80 ≤ Z ≤ 0)
(d) P(−2.80 ≤ Z ≤ 2.80)
(e) P(Z ≤ 1.14)
(f) P(−1.45 ≤ Z)
(g) P(−1.80 ≤ Z ≤ 2.00)
(h) P(1.14 ≤ Z ≤ 2.50)
(i) P(1.80 ≤ Z)
(j) P(|Z| ≤ 2.50)

Let Z be a standard normal random variable and
calculate the following probabilities, drawing pictures wherever
appropriate. (Round your answers to four decimal places.)
a) P(0 <= Z <= 2.73)
b) P(Z <= 1.63)
c) P(-1.05 <= Z)
d) P(|Z| <= 2.5)
How do you do this or is there an easy way you can do this in
excel?

Let z denote a standard normal random variable.
a. Find P(z > 1.48).
b. Find P(-0.44 < z < 2.68).
c. Determine the value of which
satisfies P(z > z. ) = 0.7995. d. Find P(z <
–0.87).

Let Z be a standard normal random variable and
calculate the following probabilities, drawing pictures wherever
appropriate. (Round your answers to four decimal
places.)
(e) P(Z ≤ 1.43)
(h) P(1.43 ≤ Z ≤
2.50)

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