What is the probability that a student will be over 60 inches tall? (assume the

Question

What is the probability that a student will be over 60 inches tall? (assume the data are normally distributed with the population standard deviation=4.25)

What is the probability that a student received a grade of 12 or above on the first experiment? The second experiment? The third Experiment? (assume the data are normally distributed with the population standard deviation=1.53)

Conclusion: use the results above to discuss the progress of students in terms of their grades among the three experiments.

***please show all solutions with formulas in excel***

Student ATT HW Experement1 Experement2 Experement 3 Experement Total Height Gender Class Major 1 8 16 11.6 9.9 10.5 32.0 66 MALE SO OTH 2 10 17 12.7 9.6 10.2 32.5 63 FEMALE JR MGMK 3 10 19 14.3 8.7 10.7 33.7 59 FEMALE SO OTH 4 7 15 10.1 8.7 10.3 29.1 66 MALE JR ACCT 5 8 11 11.6 8.3 11.4 31.2 74 MALE JR MGMT 6 8 17 13.1 8.7 10.7 32.6 65 FEMALE JR MGMT 8 10 17 13.9 9.2 9.1 32.1 58 FEMALE JR ACCT 10 10 13 14.4 9.2 11.1 34.6 65 FEMALE JR MKTG 12 8 14 16.0 10.4 9.7 36.1 78 MALE JR FINC 13 8 18 17.2 9.3 10.4 36.9 72 MALE JR MGMT 14 10 22 14.3 12.6 13.1 39.9 65 FEMALE JR FINC 15 8 17 17.0 8.1 10.1 35.1 66 FEMALE JR MKTG 16 10 21 18.7 11.7 11.9 42.3 67 MALE SO MKTG 18 10 20 15.8 12.3 11.7 39.8 68 MALE SO OTH 19 10 15 12.3 8.6 9.3 30.1 73 MALE SO FINC 20 10 19 12.5 10.7 11.6 34.7 75 MALE JR MGMT 21 8 14 13.4 9.2 9.6 32.1 73 MALE JR MGMT 22 8 17 14.0 11.0 12.3 37.2 73 MALE JR MGMT 23 10 20 19.0 12.5 14.0 45.4 71 MALE JR ACCT 24 10 21 16.6 8.7 13.4 38.7 73 MALE JR MGMT 25 10 18 17.5 12.9 11.9 42.3 76 MALE SO OTH 26 10 9 17.9 7.4 9.6 34.8 77 MALE JR MGMT 27 8 19 14.4 10.2 12.0 36.6 77 MALE JR MGMK 28 8 17 13.4 11.0 9.2 33.5 60 FEMALE JR MKTG 29 6 6 10.2 4.2 7.5 21.9 71 MALE SR MKTG 30 10 15 13.2 8.3 10.4 31.8 72 MALE JR MGMT 31 10 18 14.5 9.9 10.7 35.0 65 FEMALE JR MKTG 32 10 15 12.0 9.6 11.0 32.6 73 MALE JR MKTG

Explanation / Answer

1)

Average height (mu) = 69.32

P(z>-2.19) = .0143

Hence P(x > 60) = P(z > -2.19) = [total area] - [area to the left of -2.19]

P = 1 - .0143 = .9857

2) Experiment 1

Average height of experiment 1 is 14.34

Hence P(x > 12) = P(z > -0.55) = [total area] - [area to the left of -0.55]

= 1 - 0.2946

= 0.7054

Experment 2

Hence P(x > 12) = P(z > 0.55) = [total area] - [area to the left of 0.55]

= 0.2946

Expermient #

Hence P(x > 12) = P(z > 0.28) = [total area] - [area to the left of 0.28]

= 0.6103

The peobability of scoring more than 12 is high in Experiment 1 and experiment 3. Whereas it's very low in experiment 2.

Given, mu       = 69.32 X           = 60 n          = 28 Standard Deviation (SD)= 4.25 Standard Error (S.E) = 0.80 Z    = ( X - mu) / SD Z    = -2.19

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