Question

Z is a standard normal random variable, then k is ...

a. P(Z < k) = 0.92

b. P(Z > k) = 0.72

c. P(Z ≤ k) = 0.26

d. (A-Grade) P(−1 < Z < k) = 0.60

e. (A-Grade) P(k < Z < 1.7) = 0.57

f. (A-Grade) P(Z = k) = 0.00

Answer 1

Here Z is standard normal random variable i.e, Z~N(0,1)

a) P(Z<k)=0.92

Solution::

Here we need to find the value of k

P(Z<k)=0.92

From standard normal table we get

k=1.405

b) P(Z>k)=0.72

Solution::

Here we need to find the value of k

P(Z>k)=0.72

From standard normal table we get the value of

k=-0.583

c) ​​

Solution::

Here we need to find the value of k

From the standard normal table we get the value of

k=-0.643

d) P(-1<Z<k)=0.60

Solution::

Here we need to find the value of k

P(-1<Z<k)=0.60

[From standard normal table we get the value of ]

From standard normal table we get the value of

k=0.703

e) P(k<Z<1.7)=0.57

Solution::

Here we need to find the value of k

P(k<Z<1.7)=0.57

[ From the standard normal table we get ]

From the standard normal table we get the value of

k=-0.292

f) P(Z=k)=0.00

Solution::

Here we need to find the value of k

Here, Z~N(0,1) i.e, Z is a continuous random variable

So, in any particular point probability is zero

So, for any real value of k, P(Z=k)=0.00

[ since the range of the random variable Z is whole real line i.e, ]