The Mean height of students in a large MSU classroom of 121 students is 5\' 8\"

Question

The Mean height of students in a large MSU classroom of 121 students is 5' 8" (68 inches) with a standard deviation of 12, with a distribution that could be viewed as being normal. (That is, mu = 68 and sigma = 12). 1. If ONE student is selected at random, a. What is the probability that he/she be shorter than 62-inches tall? Calculate and demonstrate. b. What is the probability that this student's height is between 66 inches and 72 inches? Calculate and demonstrate. 2. If a group of 36 students are randomly selected, what is the probability that the mean of this random sample is more than 70 inches x^bar =70)? Calculate and demonstrate. 1. Suppose this class is viewed as a random sample of the students at MSU (In this case you know that we are saying: x^bar = 68 and S = 12) a. Construct a 98% confidence interval about the value of the Mean height of MSU students based on this sample of 121 students in this class. b. How does your answer to the part a (previous question), change if this sample had 25 students, instead of 121 students? In other words, construct another 98% Confidence Interval about the value of the University mean height, based on this new sample size and the same mean and standard deviation for this sample as that before. 2. Now, assume the University officials have claimed the mean height of the students for the whole university is 71 inches (mu = 71). Do you reject their claim with 95% certainty, or do you fail to reject (In other words, test a hypothesis regarding the validity of the claim by the university officials ((mu =71) using the data for the initial sample of 121 students in that class with mean of 68 and standard deviation of 12). Demonstrate and calculate.

Explanation / Answer

PARTA:

normal distribution with

mean=68 sd=12

(a)

shorter than 62

P(X<62)

convert x to z

z=x-mean/sd/sqrt(n)

=62-68/12/sqrt(1)

=-6/12/1

=-1/2

z =-0.5

need to find P(z<-0.5)

P(z<-0.5)=1-P(Z<0.5)

=1-0.6915

=0.3085

ANSWER 30.85% CHANCE IS THERE THAT HE IS SHORTER THAN 62 INCH

Solutionb:

P(66<X<72) to be found

P(66-68/12<z<72-68/12)

P(-0.17<z<0.33)

=P(Z<0.33)-P(Z<-0.17)

=0.6293-0.4325

=0.1968

answer 0.1968

19.68% probability that the students height is between 66 inches and 72 inches

  

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