# Use the following Normal Distribution table to calculate the area under the Normal Curve (Shaded area in the Figure) when Z=1.3 and H=0.05; Assume tha

Use the following Normal Distribution table to calculate the area under the Normal Curve (Shaded area in the Figure) when $$Z=1.3$$ and $$H=0.05$$;
Assume that you do not have vales of the area beyond $$z=1.2$$ in the table; i.e. you may need to use the extrapolation.
Check your calculated value and compare with the values in the table $$[for\ z=1.3\ and\ H=0.05]$$.
Calculate your percentage of error in the estimation.
How do I solve this problem using extrapolation?
$$\begin{array}{|c|c|}\hline Z+H & Prob. & Extrapolation \\ \hline 1.20000 & 0.38490 & Differences \\ \hline 1.21000 & 0.38690 & 0.00200 \\ \hline 1.22000 & 0.38880 & 0.00190 \\ \hline 1.23000 & 0.39070 & 0.00190 \\ \hline 1.24000 & 0.39250 & 0.00180 \\ \hline 1.25000 & 0.39440 & 0.00190 \\ \hline 1.26000 & 0.39620 & 0.00180 \\ \hline 1.27000 & 0.39800 & 0.00180 \\ \hline 1.28000 & 0.39970 & 0.00170 \\ \hline 1.29000 & 0.40150 & 0.00180 \\ \hline 1.30000 & 0.40320 & 0.00170 \\ \hline 1.31000 & 0.40490 & 0.00170 \\ \hline 1.32000 & 0.40660 & 0.00170 \\ \hline 1.33000 & 0.40830 & 0.00170 \\ \hline 1.34000 & 0.41010 & 0.00180 \\ \hline 1.35000 & 0.41190 & 0.00180 \\ \hline \end{array}$$

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

curwyrm

Step 1
Here, we have to use the technique of extrapolation for finding the probability for the $$z = 1.35 (Z=1.3\ \&\ H = 0.05).$$ Let us see this discuss this question step by step given as below:
For using extrapolation we need the values for the row of $$z = 1.2$$
The values for the row z and columns H are given as below:
By using the extrapolation, we get
$$\begin{array}{|c|c|}\hline Z+H & Prob. & Extrapolation \\ \hline 1.20000 & 0.38490 & Differences \\ \hline 1.21000 & 0.38690 & 0.00200 \\ \hline 1.22000 & 0.38880 & 0.00190 \\ \hline 1.23000 & 0.39070 & 0.00190 \\ \hline 1.24000 & 0.39250 & 0.00180 \\ \hline 1.25000 & 0.39440 & 0.00190 \\ \hline 1.26000 & 0.39620 & 0.00180 \\ \hline 1.27000 & 0.39800 & 0.00180 \\ \hline 1.28000 & 0.39970 & 0.00170 \\ \hline 1.29000 & 0.40150 & 0.00180 \\ \hline 1.30000 & 0.40320 & 0.00170 \\ \hline 1.31000 & 0.40490 & 0.00170 \\ \hline 1.32000 & 0.40660 & 0.00170 \\ \hline 1.33000 & 0.40830 & 0.00170 \\ \hline 1.34000 & 0.41010 & 0.00180 \\ \hline 1.35000 & 0.41190 & 0.00180 \\ \hline \end{array}$$
We get $$P(Z<1.35) = 0.4119$$ by using the extrapolation however the probability by using the table is given as 0.4115.
This means there is a very small difference observed due extrapolation.