Question

1. The marks in a university statistics course are normally distributed with a mean of 68% and a standard deviation of 6%.

1. Sketch the normal distribution for the course. Label the scale on the horizontal axis.

2. Calculate the z-score for a student with a mark of 79%, and explain what it means.   

3. Calculate the probabilities for a student to have the following grades:   

               (i) Greater than 60% (ii) Between 70% and 80%

2. The mid tirm grades had a course mean of 80% and a standard deviation of 6%. Mrs. Lohan wishes to bell curve the mean to 84% with a standard deviation of 10%. If you received a mark of 85%, what is your new mark?

3. The heights of students of Woodlands Summer School have a mean of 148 cm with a standard deviation of 10cm. In what range of heights would you have to be in to be in the middle 80% of heights (meaning 10% above, 10% below)   

Answer 1

let variable the (1) random х be the situation which represent *~ N(0.08, 0.00 if be will » The graph 0.16 0.92 0:14 a 62 0.68 0.5 o s6 (2) Cliven X .79. score z store the 2- 0.79-0.68 0.06 1.8333 The point 1.8333 represents the marke the standard normal curve in 790%. (3) P(x 2.6

Р s -1.333 3.) z 0.9088 (ii) pl 0.70 2 x E o so 0.80) 0.30-0.68 § 2 < 0.80-0.68 0.06 0.06 0.3333 ****** 5250) z 0.3467 random variable Let be.. the mid term. which represent x~ N (0.80, 0.00) marks . receive 0,85 Givero, we 2 z 0.85 1 is calculated of nzo.85 The 3-score as 21 - M N 8

Z 0.85 -0.80 0.06 0.83 3 3 of Lindsley. 7 Let the mark represents Yu NC 0.841 'the a-score Now, changing 0.10) to new scale is 6.10t 0.84 0.8 3 3 3 X 0.9233 obtained 0.8 marles 92.33°16 new way the . in the random N the Let represents of the height of variable students. zu all 148, 10). Clearly, normal the standard denote Let 3 A variable

P( 85 3 <3 20.10, 3') Using normal table, the 10th percentili is -1.2815. -- goth Similarly, the pereenhle of is 12815 standard. normal distribution i Using Using ' can easily +1.2815) find out the a-values 2 The n-values ale to*155 +1.2815 x 10+ 148 135.18 160.8155 should be in Therefore, the height should and 160.816 , 135.185 between,