for a standard normal distribution, find the approximate value of p(z≥ - 1.25). use the portion of the…

for a standard normal distribution, find the approximate value of p(z≥ - 1.25). use the portion of the standard normal table below to help answer the question.\n| z | probability |\n| ---- | ---- |\n| 0.00 | 0.5000 |\n| 0.25 | 0.5987 |\n| 1.00 | 0.8413 |\n| 1.25 | 0.8944 |\n| 1.50 | 0.9332 |\n| 1.75 | 0.9599 |\n11%\n39%\n61%\n89%

for a standard normal distribution, find the approximate value of p(z≥ - 1.25). use the portion of the standard normal table below to help answer the question.\n| z | probability |\n| ---- | ---- |\n| 0.00 | 0.5000 |\n| 0.25 | 0.5987 |\n| 1.00 | 0.8413 |\n| 1.25 | 0.8944 |\n| 1.50 | 0.9332 |\n| 1.75 | 0.9599 |\n11%\n39%\n61%\n89%

Answer

Answer:

D. 89%

Explanation:

Step1: Recall property of normal distribution

The total area under the standard - normal curve is 1. Also, $P(Z\geq - 1.25)=1 - P(Z\lt - 1.25)$.

Step2: Use symmetry of normal distribution

Since the standard normal distribution is symmetric about $z = 0$, $P(Z\lt - 1.25)=P(Z\gt1.25)$. And $P(Z\gt1.25)=1 - P(Z\leq1.25)$.

Step3: Find $P(Z\leq1.25)$ from the table

From the given standard - normal table, $P(Z\leq1.25)=0.8944\approx0.89$.

Step4: Calculate $P(Z\geq - 1.25)$

$P(Z\geq - 1.25)=1 - P(Z\lt - 1.25)=1-(1 - P(Z\leq1.25))=P(Z\leq1.25)\approx0.89 = 89%$.