Answer:
Minimum score = 629
Step-by-step explanation:
In order to solve this problem we must find the x-value for which the area on the topmost corner of the bell curve is 0.08 or 8% (See attached picture). In order to do so, we need to make use of a normal distribution table. There are different tables for normal distribution, the one I have gives us the areas between the mean and the value we are analyzing, so we need to first find what the area between the mean and the value we need is. We can do so by subtracting:
0.5-0.08=0.42
Now, depending on how accurate your table is, you might need to do some interpolation. The z-score I got from my table for an area of 0.42 is:
z=1.405
so I can use this value with the z-score formula to build an equation:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
where z is the z-score, x is the value we need μ is the mean and σ is the standard deviation. So when substituting the values we get:
[tex]1.405=\frac{x-455}{124}[/tex]
when solving for x we get that:
x=124(1.405)+455
so
x=629.22
So the minimum score required for the job offer is 629.
Answer:
Minimum Score required for top 8% students to secure the job offer will be 629.
Step-by-step explanation:
Since university plans to offer tutoring jobs to students whose scores are in top 8% means that there will be 92% of students failing to secure tutoring jobs.
The corresponding Z-score can be read from normal distribution table,
invNorm(0.92)=1.4051
Now Solving it for "X" using the follwing equation,
X=Z*S+U
Where "S" is Standard Deviation, "U" is the mean in above equation with Z=1.4051,
Hence Solving the above equation for the minimum score "X"
X=1.4051*124+455
X=174.2324+455
X=629.2324
Rounding it off to the nearest complete number,
X=629
Hence the Minimum Score required for top 8% students to secure the job offer will be 629.
