Question

SLMDOLS Benchmark Assessment 28 (2 points) Which measure of central tendency is preferred for this data set? Why? 56, 40, 58, 41, 53, 55, 46, 45, 60, 47, 51, 56, 43, 44, 48, 54, 60, 50, 57, 40, 46, 47, 49, 59, 44, 50, 42, 53, 195,41 (2 points) A group of students conducts a survey of 40 classmates about the number of hours they sleep in a typical night. The results are summarized below in a frequency distribution. 30 Frequency Hours of Sleep 0-2 3-5 6-8 9-11 12+ 15 Construct a histogram for the frequency distribution above. Estimate the position of the mean, median and mode. (2 points) The amount of rent paid per month for apartments in a neighborhood is normally distributed. The mean rent is $675 with a standard deviation of $50. Estimate what percent of the apartment residents pay from $625 to $775 per month. 29

Answer 1

Ans:

28)As above data is skewed and there is one outlier i.e. 195,we will preferred median as the measure of central tendency.

29)

Mean=675

standard deviation=50

625 is one standard deviation below the mean and 775 is 2 standard deviations above the mean.

z(625)=(625-675)/50=-1

z(775)=(775-675)/50=2

P(-1<z<2)=P(z<2)-P(z<-1)

=0.9773-0.1587=0.8186 or 81.86%

30)

Mode will lie between 3-5,as it has highest frequency.

Median will also be between 3-5,as median is the data value which divides the data into two equal halves.

hrs of sleep	midpoint	frequency	m*f
0-2	1	9	9
3-5	4	15	60
6-8	7	8	56
9-11	10	5	50
12+	13	3	39
Total		40	214

Mean=214/40=5.35