Question

How do you find the Z – score such that the area under the standard normal curve to the left is 0.93?

Answer 1

We solve this problem by using the table of Z – scores that correspond to the area under the graph. The standard table of conversions for the area to Z – scores and vice versa for area less than 0.5 is given below. We take the values near to 0.45 from the table and find the Z – score for 0.45. To find the Z – scores we need to check the horizontal and vertical values and add them to get the Z – score corresponding to that area. We need to note that the Z – scores for areas greater than 0.5 are positive.

Complete step-by-step solution:
We use the interpolation theorems to find the required value that are,
The part from $b$ the point $a$ lies considering $b,c$ are nearest less than and greater than values respectively to $a$ is given as $\dfrac{a-b}{c-b}$

We are asked to find the Z – score that corresponds to the area of 0.93.
Let us assume that the required Z – score as $'z'$
Here, we can see that the table used for standard conversions of area under graph to Z – score does not include 0.93.
So, let us take the two values of the area nearest to 0.93 before and after 0.93 from the table.
The least nearest area given in the graph is 0.9292. Here, we can see that the horizontal and vertical values of Z – score that represent 0.9292 are 1.4 and 0.07.
We know that we need to add horizontal and vertical values of Z – scores to the Z – score corresponding to that respective area.
We know that for the area greater than 0.5 the Z – score will be positive.
So, we can say that the Z – score for 0.9292 is given as,
$\Rightarrow {{z}_{1}}=1.47$
The nearest area given in the graph is 0.9306. Here, we can see that the horizontal and vertical values of Z – score that represent 0.9306 are 1.4 and 0.08.
So, we can say that the Z – score for 0.9306 is given as,
$\Rightarrow {{z}_{2}}=1.48$
Now, let us find the ratio at which the area 0.93 lies from 0.9292 in between 0.9292, 0.9306.
We know that the ratio for which the point $a$ lies from $b$ considering $b,c$ are nearest less than and greater than values respectively to $a$ is given as $\dfrac{a-b}{c-b}$
By using this theorem we get the ratio at which 0.45 lies from 0.4483 in between 0.4483, 0.4522 is
$\Rightarrow \dfrac{0.9300-0.9292}{0.9306-0.9292}=\dfrac{0.0008}{0.0014}=\dfrac{8}{14}$
Here, we can see that area 0.93 is $\dfrac{8}{14}$ apart after from 0.9292 in between 0.9292, 0.9306.
So, we can say that the value of $z$ is $\dfrac{8}{14}$ part from ${{z}_{1}}$ in between ${{z}_{1}},{{z}_{2}}$
Now, we can rewrite the above theorem as, if $x$ is $f$ part from $y$ considering $y,k$ is given as,
$\Rightarrow x=y+f\left( z-y \right)$
By using this result we get the required value of Z – score as,
$\begin{aligned} & \Rightarrow z=1.47+\dfrac{8}{14}\left( 1.48-1.47 \right) \\\ & \Rightarrow z=1.47+0.0057 \\\ & \Rightarrow z=1.4757 \\\ \end{aligned}$
$\therefore$ Therefore, we can conclude that the Z – score corresponding to area 0.93 is given as $1.4757$

Note: We need to note that if the area given is greater than 0.5 then the Z – score should be always positive. So, we need positive Z – scores from the table corresponding to the assumed areas.
Also we need to note that the interpolation theorem is used to find an approximate value corresponding to one area if we consider some interval. Here, as we need to find Z – score corresponding to 0.93 we take the interval as $\left[ 0.9292,0.9306 \right]$ which are nearest less than and greater than values of 0.93 from the table.