which of the following probabilities is equal to approximately 0.2957? use the portion of the standard…

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.\n| z | probability |\n|----|----| \n| 0.00 | 0.5000 |\n| 0.25 | 0.5987 |\n| 0.50 | 0.6915 |\n| 0.75 | 0.7734 |\n| 1.00 | 0.8413 |\n| 1.25 | 0.8944 |\n| 1.50 | 0.9332 |\n| 1.75 | 0.9599 |\np(-1.25≤z≤0.25)\np(-1.25≤z≤0.75)\np(0.25≤z≤1.25)\np(0.75≤z≤1.25)

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.\n| z | probability |\n|----|----| \n| 0.00 | 0.5000 |\n| 0.25 | 0.5987 |\n| 0.50 | 0.6915 |\n| 0.75 | 0.7734 |\n| 1.00 | 0.8413 |\n| 1.25 | 0.8944 |\n| 1.50 | 0.9332 |\n| 1.75 | 0.9599 |\np(-1.25≤z≤0.25)\np(-1.25≤z≤0.75)\np(0.25≤z≤1.25)\np(0.75≤z≤1.25)

Answer

Explanation:

Step1: Recall the property of standard - normal distribution

The probability $P(a\leq Z\leq b)=\Phi(b)-\Phi(a)$, where $\Phi(z)$ is the cumulative - distribution function of the standard normal distribution and values can be read from the standard normal table.

Step2: Check option A

For $P(- 1.25\leq Z\leq0.25)$, $\Phi(0.25) = 0.5987$ and $\Phi(-1.25)=1 - \Phi(1.25)=1 - 0.8944 = 0.1056$. Then $P(-1.25\leq Z\leq0.25)=0.5987-0.1056 = 0.4931$.

Step3: Check option B

For $P(-1.25\leq Z\leq0.75)$, $\Phi(0.75)=0.7734$ and $\Phi(-1.25)=1 - \Phi(1.25)=1 - 0.8944 = 0.1056$. Then $P(-1.25\leq Z\leq0.75)=0.7734 - 0.1056=0.6678$.

Step4: Check option C

For $P(0.25\leq Z\leq1.25)$, $\Phi(1.25)=0.8944$ and $\Phi(0.25)=0.5987$. Then $P(0.25\leq Z\leq1.25)=0.8944 - 0.5987=0.2957$.

Step5: Check option D

For $P(0.75\leq Z\leq1.25)$, $\Phi(1.25)=0.8944$ and $\Phi(0.75)=0.7734$. Then $P(0.75\leq Z\leq1.25)=0.8944 - 0.7734 = 0.121$.

Answer:

$P(0.25\leq Z\leq1.25)$