#4. On a qualifying exam, Carly scored at the 34th percentile and Joseph scored
ID: 3071488 • Letter: #
Question
#4. On a qualifying exam, Carly scored at the 34th percentile and Joseph scored at the 68th percentile. A classmate explained that Joseph's score was twice as high as Carly's score. Is this the correct retation of percentiles? Justify your answer. #5. A random sample of 75 college students were asked how many cans of soft drinks they typically consume on a given day. That number was multiplied by 10 to give a daily amount of sugar (tsp) from drinking soft drinks. The following statistics were calculated Min - 10 Max-61 Lower quartile 25 Median 31 Mean-31.4 Upper quartile 38 n 75 s 11.66 (a) I f these statistics were maintained for school of 3200 studeats, how many students consume between 31 and 38 tsps of sugar in a day? (b) Are there any outliers in this set of data? Justify your answer If so, what type? 6. A batch of 12 numbers had an arithmetic average of 117. The second largest number was 174 but it was determined that the value had been accidentaly transposed and should be 147 (a) By how much will the average change? (b) By how much will the median change #7. Individuals are to be chosen from a pool of twenty engineers with the city fire department to be people will remain unassigned ) How many ways can the assignment be made? (An expression is sufficient) 34, P(D)-53, and penD) 2, #8. Let C and D be two events with P(C) What is P(CnD)? #9. Two fair dice are rolled. Consider the events A: the sum of the two dice equals 11 B: the sum of the two dice equals 7 C: at least one of the dice shows a 6 (a) P(AC)- (b) P(BC)- (c) Are B and C independent? ExplainExplanation / Answer
#6.
a)
n = 12, mean = 117
sum = 12*117 = 1404
1404 - 174 = 1230
1230 + 147 = 1377
new mean = 1377/12 = 114.75
change in mean = 117 - 114.5 = 2.5
b)
There will be no change in the median.
Because median is the average of 6th and 7th number in the list of numbers arranged in the ascending order.
#7.
20 engineers
5 engineers for district I can be selected in 20C5 ways
3 engineers for district II can be selected in 15C3
4 engineers for district III can be seleced in 12C4
Required number of ways = 20C5*15C3*12C4 = 3491888400
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