Question

The following are the IQ scores of 60 students from a university. The histogram of the data is symmetric. 79 79 81 82 82 84 84 84 85 85 86 86 87 89 89 89 90 90 90 91 91 92 92 93 94 95 95 96 98 99 99 102 102 102 103 104 104 104 105 106 106 106 108 109 109 110 111 111 113 113 113 115 117 117 120 121 123 124 125 130

a. For this data, the 5-number statistics is (i) 79 89 99 109.5 130 (ii) 79 89.5 99 109 130 (iii) 79 89 99.5 110 130 (iv) none of the preceding

b. The mean and the standard deviation are (i) 99.82 and 13.23 (ii) 99.6 and 175.07 (iii) 99.82 and 13.12 (iv) none of the preceding

c. The z-score of the score 111 is (i) 0.8521 (ii) 0.8450 (iii) 0.8616 (iv) none of the preceding

d. If the z-score of a student is 0.6183, then the student’s IQ score is (i) 111 (ii) 108 (iii) 110 (iv) none of the preceding

e. The interval (¯ x−s, ¯ x+s) is approximately (i) [87, 113] (ii) [89, 115] (iii) [85, 111] (iv) none of the preceding

f. According to the empirical rule (i) about 68% of the scores fall in ¯ x±s, but the actual percentage of scores in the interval is 65. (ii) about 95% of the scores fall in ¯ x±s, but the actual percentage of scores in the interval is 65, which is much smaller than 95. (iii) about 50% of the scores fall in ¯ x±s, but the actual percentage of scores in the interval is 49, which is close to 50. (iv) about 16% of the scores are greater ¯ x+s, but the actual percentage of scores greater than ¯ x+s is 12.

Answer 1

The 60 students summary are

79 79 81 82 82 84 84 84 85 85 86 86 87 89 89 89 90 90 90 91 91 92 92 93 94 95 95 96 98 99 99 102 102 102 103 104 104 104 105 106 106 106 108 109 109 110 111 111 113 113 113 115 117 117 120 121 123 124 125 130

a) The five point summary are

(i) 79 89 99 109.5 130

b) Mean calculated as

and standard deviation calculated as

sample standard deviation as

is standard deviaton is considered

(iii) 99.82 and 13.12

if Sample standrd devition is considered

(i) 99.82 and 13.23

c). The Z score for 111 calculated as if standard deviation

$Z=\frac{X-\bar{x}}{\sigma}= \frac{111-99.82}{13.12}=0.8521$

(i) 0.8521

if sample standard deviation considered

$Z=\frac{X-\bar{x}}{\sigma}= \frac{111-99.82}{13.23}=0.8450$

(ii) 0.8450

d).et the Z score =0.6183 hence IQ score as

$Z=\frac{X-\bar{x}}{\sigma}= \frac{X-99.82}{13.12}=0.6183$

hence X,, (ii) 108

e) According to interval (¯ x−s, ¯ x+s) the value is approximately

(i) [87, 113]

f) According to the empirical rule (i) about 68% of the scores fall in ¯ x±s, but the actual percentage of scores in the interval is 65.