Math 1065 Quantitative Methods In Assessment Answers

Questions:

Answer:

it can be inferred that the percentage of people who took a nap and who did not take a nap in the age group of 80+ is approximately equal. For other groups, 18 – 29, 30 – 49, 40 – 69 and 70 – 79 the percentage of people who did not take a nap is more than the percentage of people who did take a nap.

it can be inferred that percentage of people who were pretty happy and very happy when they did not take a nap was more than when they did take a nap.

Table 1: Distribution Age and Nap 

Rows: Age   Columns: Nap

Cell Contents
      Count
      % of Row

From the above table it is found that 52.38% of 80+ age citizens take a nap. 67.47% of people in the age group of 50 – 69 years’ age do not take a nap.

Table 2: Distribution mood and Nap  

Rows: Mood   Columns: Nap

Cell Contents
      Count
      % of Row

From the above table it can be inferred that of “pretty happy” citizens 73.61% did not take a nap. Similarly, “very happy” citizens 71.27% did not take a nap. On the other hand, from “Not too happy” people 53.16% did not take a nap and 46.84% did take a nap.

The histogram (figure 3) of Cold days of the two groups suggests that the distribution of cold days is positively skewed. Thus it can be inferred that the mean number of cold days is less than the median number of cold days.

From the (figure 4) there is no indication of outliers in number of cold days.

From both the histogram and boxplot it can be inferred that the data for cold days is right skewed. Since the data is not normally distributed, hence median would be the appropriate measure of central tendency. The median number of Cold days for Elderberry is 4.0 and for Placebo is 6.0 respectively. The median is used as a measure of central tendency. Thus IQR would be the ideal measure of dispersion. The IQR for cold days for Elderberry is 3.750 (6.750 – 3.000). The IQR for cold days for Placebo is 4.500 (9.50 – 4.00).

The mean number of cold days using Elderberry is 4.667 while for placebo is 6.882. The maximum number of cold days in Elderberry is 9.00 while for Placebo is 15.00. Moreover, the median number of cold days for Elderberry is 4.00 and for Placebo is 6.00. Thus it can be inferred that by using Elderberry the number of Cold days is less as compared to Placebo.   

Let X denote the mean glucose level

The mean glucose level is normally distributed with  and

The probability that the mean glucose level is P(X>140mg/dl)

Method 1: Use Standard Normal Tables

z-score for 

Thus,

From z-table it is found that P(z = -1.5) = 0.933 

Therefore, the probability that the patient has gestational diabetes = 0.067

Using MINITAB, a normal distributed random sample of size 5 is created.

The mean and standard of the sample is 128.27 and 12.02 respectively.

Method 1: Use Standard Normal Tables

z-score for 

Thus,

From z-table it is found that P(z = ) = 0.83522 

Therefore, the probability that the patient has gestational diabetes = 0.16478

Method 2: Minitab 

Method 1: Use Standard Normal Tables

Rearranging the distribution from ascending to descending:

The median value is = 133.29

The first quartile =

The third quartile =

Thus, the inter-quartile range =  

Method 2: Minitab

The third quartile value for glucose = 138.19.

The first quartile value for glucose = 115.93

Thus the Inter quartile range for the amount of glucose = 138.19-115.93 = 22.26 

Method 1: Use Standard Normal Tables

Let the mean waiting time be represented as

Let the standard deviation of the waiting time be represented as

Thus the probability that the mean waiting time is less than 15 minutes is represented as

Further, the probability that the mean waiting time is more than 60 minutes is represented as

From calculations it is found that for P = 0.25, z = -0.6745

For P(z<-06745), z = -0.6745

From calculations it is found that for P = 0.05, z = -1.6449

For P(z>-1.6449), z = 1.6449

Thus,                                    (i)

and                                          (ii)