Question

Which of the following probabilities is equal to approximately 0. 2957? Use the portion of the standard normal table below to help answer the question. Z Probability 0. 00 0. 5000 0. 25 0. 5987 0. 50 0. 6915 0. 75 0. 7734 1. 00 0. 8413 1. 25 0. 8944 1. 50 0. 9332 1. 75 0. 9599 P (negative 1. 25 less-than-or-equal-to z less-than-or-equal-to 0. 25) P (negative 1. 25 less-than-or-equal-to z less-than-or-equal-to 0. 75) P (0. 25 less-than-or-equal-to z less-than-or-equal-to 1. 25) P (0. 75 less-than-or-equal-to z less-than-or-equal-to 1. 25).

Answer 1

Probability of an event is the measure of its chance of occurrence. The event out of the listed events whose probability is 0.2957 is given by : Option C: $P(0.25 \leq Z \leq 1.25)$

How to get the z scores?

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z-score.

If we have

$X \sim N(\mu, \sigma)$

(X is following normal distribution with mean $\mu$ and standard deviation $\sigma$ )

then it can be converted to standard normal distribution as

$Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)$

(Know the fact that in continuous distribution, probability of a single point is 0, so we can write

$P(Z \leq z) = P(Z < z) )$

Also, know that if we look for Z = z in z tables, the p-value we get is

$P(Z \leq z) = \rm p \: value$

Using the z-table, we get the needed probabilities as:

Case 1:

$P(-1.25 \leq Z \leq 0.25) = P(Z \leq 0.25) - P(Z \leq -1.25) \approx 0.5987 - 0.1056 = 0.4931$

Case 2:

$P(-1.25 \leq Z \leq 0.75) = P(Z \leq 0.75) - P(Z \leq -1.25) \approx 0.7734- 0.1056=0.6678$

Case 3:

$P(0.25 \leq Z \leq 1.25) = P(Z \leq 1.25) - P(Z \leq 0.25) \approx 0.8944 - 0.5987=0.2957$

Case 4:

$P(0.75 \leq Z \leq 1.25) = P(Z \leq 1.25) - P(Z \leq 0.75) \approx 0.8944 - 0.7734 =0.1210$

Thus, the event out of the listed events whose probability is 0.2957 is given by : Option C: $P(0.25 \leq Z \leq 1.25)$

Learn more about z-scores here:

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