Suppose GRE Verbal scores are normally distributed with a mean of 455 and a standard deviation of 124. A university plans to offer tutoring jobs to students whose scores are in the top 8%. What is the minimum score required for the job offer? Round your answer to the nearest whole number, if necessary.

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:

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:

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.

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752