To get the z-value of the scores of the four students, we are going to use the formula for standard score or z-score. It is score minus the mean score, then divided by standard deviation.
z= Score (X)-Mean / SD
To find the z-value of each score, we have to use a Z table. Using the z-score, we are to look first at the y-axis of the table which will highlight the first two digits of the z-score. Then, the x-axis for the second decimal place of the z-score.
You can use this as reference for the z-table: http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf
Mean= 500SD= 100Scores= 560, 450, 640, 530 For the student who scored 560,z= X-Mean / SDz= 560-500 / 100z= 60 / 100z= 0.6 The score is 0.6 standard deviation above the mean. The z-value is 0.7257 or 72.57%. For the student who scored 450,z= X-Mean / SDz= 450-500 / 100z= -50 / 100z= -0.5 The score is -0.5 standard deviation above the mean. The z-value is 0.3085 or 30.85%.
For the student who scored 640,z= X-Mean / SDz= 640-500 / 100z= 140 / 100z= 1.4 The score is 1.4 standard deviation above the mean. The z-value is 0.9192 or 91.92%. For the student who scored 530,z= X-Mean / SDz= 530-500 / 100z= 30 / 100z= 0.3 The score is 0.3 standard deviation above the mean. The z-value is 0.6179 or 61.79%.
The correct answer to the question: Suppose there is a mutation in a laminin-binding integrin gene that causes a loss of function in the cytoplasmic domains of the integrin. Which one of the following would you expect to observe as a result of this mutation?, would be, C) The strength of tissues would be decreased because the integrin could no longer associate with intermediate filaments in hemidesmosomes.
