Question

Use Table A to find the proportion of observations from the standard Normal distribution that satisfies each of the following statements. In each case, sketch a standard Normal curve and shade the area under the curve that is the answer to the question. Use your calculator or the Normal Curve applet to check your answers.

More Table A practice

(a) $z$is between −$1.33$and $1.65$

(b) $z$is between $0.50$and$1.79$

Short answer

From the given information

a) The area between $z=-1.33\&z=1.65$is$0.9505-0.0918=0.8587$.

b) The area between $z=0.50\&z=1.79$is $0.9633-0.6915=0.2718$.

Step-by-step solution

Part (a) Step 1: Given Information 

It is given in the question that, $z$is between −$1.33$and $1.65$

Part (a) Step 2: Explanation

The below Standard Normal probabilities table is a table of areas under the standard Normal Curve. The table entry for each value $z$is the area under the curve to the left of $z$.

Part (a) Step 3: Graphical Representation

shows the graph and table,

Part (a) Step 4: Explanation 

a) From the Standard Normal probabilities table, the area to the left of $z=1.65$is $0.9505$, and the area to the left of $z=-1.33$is $0.0918$.

The area between $z=-1.33$and $z=1.65$is the area to the left of $1.65$minus the area to the left of $-1.33$.

Therefore, the area between $z=-1.33$& $z=1.65$is $0.9505-0.0918=0.8587$. The result is shown in the below graph.

Part (b) Step 1: Given Information

It is given in the question that, $z$is between $0.50$and $1.79$

Part (b) Step 2: Explanation

(b) From the Standard Normal probabilities table, the area to the left of $z=1.79$is $0.9633$and the area to the left of $z=0.50$is $0.6915$.

The area between $z=0.50$and $z=1.79$is the area to the left of $1.79$minus the area to the left of $0.50$.

Therefore, the area between $z=0.50\&z=1.79$is $0.9633-0.6915=0.2718$. The result is shown in the below graph.

Therefore, the area between$z=0.50\mathrm{\&}z=1.79$is$0.9633-0.6915=0.2718$