for a standard normal distribution, find the approximate value of p(-0.78≤z≤1.16). use the portion of the…

for a standard normal distribution, find the approximate value of p(-0.78≤z≤1.16). use the portion of the standard normal table below to help answer the question. z probability 0.00 0.5000 0.16 0.5636 0.22 0.5871 0.78 0.7823 1.00 0.8413 1.16 0.8770 1.78 0.9625 2.00 0.9772 22% 66% 78% 88%

for a standard normal distribution, find the approximate value of p(-0.78≤z≤1.16). use the portion of the standard normal table below to help answer the question. z probability 0.00 0.5000 0.16 0.5636 0.22 0.5871 0.78 0.7823 1.00 0.8413 1.16 0.8770 1.78 0.9625 2.00 0.9772 22% 66% 78% 88%

Answer

Explanation:

Step1: Recall property of standard - normal distribution

$P(-0.78\leq z\leq1.16)=P(z\leq1.16)-P(z \lt - 0.78)$. Since the standard - normal distribution is symmetric about $z = 0$, $P(z\lt - 0.78)=1 - P(z\lt0.78)$.

Step2: Look up values in the table

From the table, $P(z\leq1.16) = 0.8770$ and $P(z\leq0.78)=0.7823$. Then $P(z\lt - 0.78)=1 - 0.7823=0.2177$.

Step3: Calculate the probability

$P(-0.78\leq z\leq1.16)=P(z\leq1.16)-P(z\lt - 0.78)=0.8770-(1 - 0.7823)=0.8770 - 0.2177=0.6593\approx66%$.

Answer:

66%