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**Activity 2.1 **
**1. Drug Dose **

Continuous data, i.e. you could inject 2.2ml or 2.211ml or 2.5ml

**2. Hours slept **

Continuous data, i.e. you could sleep 9hrs, 9.5 hrs or 9.75 hrs

**3. Movies watched **

Discrete data, i.e. you would have seen 5 or 6 movies not 6.5 movies

**4. Clients seen in the past month **

Discrete data, i.e. you might have had 22 clients or 25 clients

**5. Caffeine intake **

Continuous data, i.e. you could have ingested 7mg of caffeine or 7.1mg or'

6.

**Cigarettes smoked** Discrete data, i.e. you generally only smoke whole cigarettes. If asked how many you had smoked this

week you may answer 13 but not 13.5 cigarettes.

7.

**Volleyball players jersey numbers** Discrete data, i.e. you are either number 12 or 32 or 15 not 12.56 or 15.98

8.

**Gender** Discrete data, i.e. you are either male or female.

9.

**Salary ** Continuous data, i.e. you could earn any amount of rand and cents.

10.

**Televisions** Discrete data, you can watch any number of hours and minutes of television.

**Activity 2.2 **
**Frequency % Frequency **
**Table 2.2** Frequency distribution of TB outcomes

One must divide the frequency by the sample size and then multiply the answer by 100. For Item C this would mean - (19 / 50) * 100 = 38%

**Activity 2.3 **
The real upper and lower limits are found midway between the apparent

limits of neighbouring class intervals. For table 2.5, the way to establish

the RUL and RLL would be to take the distance between the apparent

upper limit of an interval and the apparent lower limit of the following class

interval. This distance would then be divided in half to provide the real

upper limit for the first interval. This same figure would represent the real

lower limit of the following interval. For the first class interval of

**table 2.5**,

** **

the RUL for the interval 80-89 would be 89.5. The RLL for the interval

would be 79.5.

The frequency table would be constructed in the same manner as table 2.5. The difference would be that the apparent limits would be altered. One would need to recalculate how many scores fall within each of the newly created class intervals. For example there are 6 scores which would fall in the first class pass interval.

**Activity 2.5 **
Firstly, you could add the three % frequency scores given. i.e. for the interval 70 - 80, 80-89 and >89.
Secondly, you could look at the cumulative % frequency for all scores below 70 and subtract it from 100. I.e. 100 - 78.75

**Activity 2.6 **
**Table 2.5** Frequency distribution table of test

marks

Midpoint of class interval 80 - 89 = 79.5 + ((89.5 - 79.5) / 2)

**Activity 2.7 **
a. Computing the median, 1st and 3rd quartile is nothing more than
computing the score associated with the 25th, 50th, and 75th percentile.

Use

**formula 2.3** to do this.

b. As in part a. simply use

**formula 2.3** to calculate score associated with

c. We are now asked to calculate the percentile rank of a score. '
be done using

**formula 2.2**
d. Two percentile rank scores must be calculated and then the smaller

**Activity 2.8 **
Computing the median, 1st and 3rd quartile is nothing more than

computing the score associated with the 25th, 50th, and 75th percentile.

Use

**formula 2.3** to do this.

Source: http://www.jutaacademic.co.za/media/support-resources/numbers/resources/tutorial2/t2activ.pdf

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