A new program implemented at the Houston College rewards outstanding teachers wi

Question

A new program implemented at the Houston College rewards outstanding teachers with the title "Distinguished Educator" and a $1,000 bonus. To receive this designation a teacher must rank in the 10% of faculty in schoo Committed to Excellence in Education. The teaching evaluation scores for all faculty in the CCEE consortium for this past year had an average of 73 with a standard deviation of 7.2. Additionally, these teaching evaluation scores will be used to identify and weed out very poor educators. The bottom 5% of teachers will be dismissed from their jobs, 2. Is that are members of the Colleges

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 73
standard Deviation ( sd )= 7.2
g.
P(X = 78) = (78-73)/7.2
= 5/7.2= 0.6944
= P ( Z <0.6944) From Standard Normal Table
= 0.7563
75.63% is the percentile for the score 78
f.
Low/Below
P ( Z < x ) = 0.05
Value of z to the cumulative probability of 0.05 from normal table is -1.6449
P( x-u/s.d < x - 73/7.2 ) = 0.05
That is, ( x - 73/7.2 ) = -1.6449
--> x = -1.6449 * 7.2 + 73 = 61.1571

UPPER/TOP
P ( Z > x ) = 0.1
Value of z to the cumulative probability of 0.1 from normal table is 1.2816
P( x-u / (s.d) > x - 73/7.2) = 0.1
That is, ( x - 73/7.2) = 1.2816
--> x = 1.2816 * 7.2+73 = 82.2272

among 100% top 10% are teachers with distingushed educator, and the bottom 5% are who dismissed educator and
we left with above 5% and below 90%
the percentage of educators left are = 100 - 5 - 10 = 85%

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