Q. 6.59 - Elementary Statistics

Question

Determine the area under the standard normal curve that lies to the left of 

a. $2.24$

b. $-1.56$

c. $0$

d. $-4$

Step-by-Step Solution

Answer

Part $\left(a\right)$: The area under the standard normal curve that lies to the left of $2.24$ is, $0.9875$.

Part $\left(b\right)$: The area under the standard normal curve that lies to the left of $-1.56$ is, $0.0594$.

Part $\left(c\right)$: The area under the standard normal curve that lies to the left of $0$ is, $0.5000$.

Part $\left(d\right)$: The area under the standard normal curve that lies to the left of $-4$ is, $0.0000$.

Part a Step 1 . Given information

We need to determine the area under the standard normal curve that lies to the left for the given values.

Part a Step src="data:image/svg+xml;base64,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" role="math" localid="1652538301910" 2 . Let us analyze the given value and find the area for it.

Since $2.24$ is positive, we use the standard normal table of positive $z$ scores. First , we go down to the left handed column labeled $z$ to $2.24$.Then going across that row to the column labeled $0.04$, we reach $0.9875$.

Therefore, the area under the standard normal curve that lies to the left of $2.24$ is, $0.9875$.

Part a Step 3 . let us draw a curve for the obtained area.

Part b Step 1 . Let us analyze the given value and find the area for it.

Since $-1.56$ is negative, we use the standard normal table of negative $z$ scores. First we go down to the right hand column labeled $z$ to $-1.5$. Then going across that row to the column labeled $0.06$, we reach $0.0594$. Therefore, the area under the standard normal curve that lies to the left of $-1.56$ is, $0.0594$.

Part b Step 2 . Let us draw a curve for the obtained area.

Part c Step 1 . Let us analyze the given value and find the area for it.

We use the standard normal table. First, we go down to the left-hand column, labeled $z$ to $0.0$. Then, going across that row to the column labeled $0.00$, we reach $0.5000$.

Therefore, the area under the standard normal curve that lies to the left of $0.00$ is, $0.5000$.

Part c Step 2 . Let us draw a curve for the obtained area.

Part d Step 1 . Let us analyze the given value and find the area for it.

Since $-4$ is negative, we use the standard normal table for negative $z$ scores. But we have no $-4.00$ in the table. So, the area under the standard normal curve that lies to the left of $-4.00$ is, $0.0000$.