Question

For a standard normal distribution, find the approximate value of p (negative 0.78 less-than-or-equal-to z less-than-or-equal-to 1.16). use the portion of the standard normal table below to help answer the question.

Answer 1

The approximate value of P(-0.78 ≤ Z ≤ 1.16) is obtained being 0.6593

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

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

For this case, we have to find:

$P(-0.78\leq Z \leq 1.16)$

It can be rewritten as:

$P(-0.78\leq Z \leq 1.16) = P(Z \leq 1.16) - P(Z < -0.78) \\P(-0.78\leq Z \leq 1.16) = P(Z \leq 1.16) - P(Z \leq -0.78)$

The p-values for Z = 1.16 and Z = -0.78 from the z-table is found as 0.8770 and 0.2177 respectively, and therefore, we get:

$P(-0.78\leq Z \leq 1.16) = P(Z \leq 1.16) - P(Z \leq -0.78)\\P(-0.78\leq Z \leq 1.16) = 0.8770 - 0.2177 = 0.6593$

Thus, the approximate value of P(-0.78 ≤ Z ≤ 1.16) is obtained being 0.6593

Learn more about z-score here:

brainly.com/question/21262765