Question

Given z is a standard normal random variable, answer questions 1 through 4.

1. p(z = 2.3) is

a. .1056.
b. 0.
c. .8944.
d. .3944.

2. p(z ≥ -1.84) is

a. .5474.
b. 0.
c. .0329.
d. .9671.

3. p(z ≤ 1.4) is

a. .0808.
b. .9192.
c. .9927.
d. 0.

4. p(.5 ≤ z ≤ 2.9) is

a. 0.
b. .6915.
c. .3066.
d. 9981.

Answer 1

Answer on Question #53370 – Math – Statistics and Probability

Given $z$ is a standard normal random variable, answer questions 1 through 4.

1. $p(z = 2.3)$ is

a. .1056.

b. 0.

c. .8944.

d. .3944.

Solution:

$p(z = 2.3) = 0$, because $z$ is a standard normal random variable.

Thus, the answer is b. 0.

2. $p(z \geq -1.84)$ is

a. .5474.

b. 0.

c. .0329.

d. .9671.

Solution:

The normal random variable of a standard normal distribution is called a standard score or a z-score. Every normal random variable X can be transformed into a z score via the following equation:

$$z = \frac{(X - \mu)}{\sigma}$$

where X is a normal random variable, $\mu$ is the mean of X, and $\sigma$ is the standard deviation of X.

In the given problem we require

$$1 - P(Z < -1.84) = 1 - \Phi(-1.84) = 1 - 0.0329 = 0.9671,$$

where $\Phi$ is the cumulative distribution function of a standard normal variable

Thus, the answer is d. .9671.

3. $p(z \leq 1.4)$ is

a. .0808.

b. .9192.

c. .9927.

d. 0.

**Solution:**

In given case if $p(z \leq 1.4)$ is 0.9192 from the normal table.

Thus, the answer is b. .9192.

4. $p(0.5 \leq z \leq 2.9)$ is

a. 0.

b. .6915.

c. .3066.

d. 9981.

**Solution:**

We have to apply the following method:

$$P(0.5 \leq z \leq 2.9) = P(Z \leq 2.9) - P(Z \leq 0.5) = \Phi(2.9) - \Phi(0.5) = 0.9981 - 0.6915$$ $$\approx 0.3066$$

Thus, the answer is c. .3066.

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