You work for a city\'s GED program (alternative to a high school diploma). You a

Question

You work for a city's GED program (alternative to a high school diploma). You are in charge in accepting applications and assigning students to classes. Each GED accommodates 25 students over the past 10 years an average of 750 people have applied each year. AS a standard deviation of 50 (assume normal distribution) calculate the statistics for the following questions and than use those statistics to write a coherent email to the education secretary to the program. HINT: Remember that number of classes and number of people are 2 different numbers. Don’t forget to convert people into classes.

a.) Last year, funding was approved for 31 classes. If funding remains the same, based on past data, what is the probability that you will have enough classes to meet the demand?

b.) There is talk of cutting funding based on past data, how many cut classes would create only a 25% chance of meeting demand?

c.) Based on past data, how many classes do we need to have a 95% probability on meeting demand?

d.) If funding were cut down to 25 classes, as one legislator proposed, what is the probability, based on past data, that the program will have enough classes to meet demand?

Explanation / Answer

a.) Funding approved for 31 classes.

Maximum of students accomodated = 31 * no of students in one class = 31*25 = 775.

Mean or average applications, mu = 750

Standard deviation , SD = 50.

Hence the probability that I will have enough classes to meet the demand = P(demand < = 775)

Z score of 775 = (x- mu )/ SD = (775-750)/ 50 = 25/50 = 0.5

Hence from the normal distribution table , P( z< = 0.5 ) = 0.691

Hence the probability that we will have enough classes to meet the demand = 0.6915 or 69.15%

b.) P ( meeting demand) = 0.25

from normal disctribution z score for P(meeting demand) = -0.6745

hence applying this in z score formula to arrive at capacity = mu-0.6745 * 50 = 750-0.6745*50 = 716.275 ~716 students

so no of classes for 716 students = 716 / 25 = 28.64 ~29 classes

Classes to be cut = 31-29 = 2 classes

c.) From normal distribution table, the z score for P(meeting demand) @95% = 1.6449

Hence capacity required = mu + 1.6449 * SD = 750 + 1.6449*50 = 832.245 ~832

Classes required = 832/25 = 33.28 ~33 classes

Classes we need to have a 95% probability on meeting demand = 33 classes

d.) Z score of 25 classes = (classes * capacity - mu)/ SD = (25*25 - 750 )/ 50 = -2.5

P(z=-2.5) from normal distribution = 0.0062 = 0.62%

The probability, based on past data, that the program will have enough classes to meet demand with 25 classes = 0.62%

Hello Secratary,

We need to have at least 33 classes to meet the demand 95 % of the times. This is arrived at since the average demand is 750 with a standard deviation of 50 over the past 10 years. With present 31 classes we will be able meet the demand only 69.15% of the times.

Please approve the required funding for 33 classes.

Yours sincerely,

ABCD

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