Statistics Problem Solving essay
← Cyberbullying in the School System  Establish Criteria for Possible Solutions → 
Statistics Problem Solving. Custom Statistics Problem Solving Essay Writing Service  Statistics Problem Solving Essay samples, help
Q.1 In a study to determine the effect of a certain gene carried out by pregnant mothers on infant weight at a later stage of their lives two samples were selected, sample (A) for mothers with the gene and sample (B) for mothers without the gene. Each sample contained 35 subjects.
Group ABirth weights of children born for mothers with the gene
27 
49 
59 
67 
75 
36 
49 
60 
68 
76 
37 
51 
60 
72 
76 
40 
52 
63 
73 
84 
42 
52 
64 
73 
102 
44 
56 
66 
74 
103 
48 
59 
66 
75 
117 
The table above shows weights in pounds for group A. Use the table to answer the following questions:
For the ungrouped data what are the value of the 25^{th }and the 75^{th} percentiles?
Answer;
P= p/100*n+1/2
Therefore, 25/100*35+1/2
0.25*35+0.5
N=9, therefore 25^{th} percentile =49
0.95*35+1/2
N=33.75=34
95^{th} percentile =103
Use the data to construct a frequency distribution and a cumulative relative frequency for grouped data? (Use classes of width of 10).
answer;
group A weights 
no of children 
cumulative weight 
27 
1 
27 
36 
1 
36 
37 
1 
37 
40 
1 
40 
42 
1 
42 
44 
1 
44 
48 
1 
48 
49 
2 
98 
51 
1 
51 
52 
2 
104 
56 
1 
56 
59 
2 
118 
60 
2 
120 
63 
1 
63 
64 
1 
64 
66 
2 
132 
67 
1 
67 
68 
1 
68 
72 
1 
72 
73 
2 
146 
74 
1 
74 
75 
2 
150 
76 
2 
152 
84 
1 
84 
102 
1 
102 
103 
1 
103 
117 
1 
117 
For the grouped data calculate the 75^{th} percentile
Answer;
75/100*27+1/2
0.75*27+0.5
=20.75=21
75^{th} percentile =74
Construct a polygon for the grouped data for Group A, and use it to describe the distribution of weights (Shape of the graph, skew ness, distribution of weights, concentration of weights, etc.. ?
Answer;
6. The table below shows data for group B weights. Plot this data after grouping it by a polygon on the same graph with the grouped data for group A, then use the resulting graph to compare between the two groups (Which has a better weight distribution? justify your answer). Note: consider an optimal weight to be between 50 and 69.
Group B has better weight distribution since its curve is significantly higher than that of group A.
Group BBirth weights of children born for mothers without the gene
30 
53 
76 
87 
97 
36 
54 
78 
87 
100 
37 
55 
79 
91 
101 
40 
62 
79 
91 
101 
42 
66 
82 
91 
102 
44 
66 
83 
92 
115 
48 
68 
85 
97 
117 
Q. 2. Dr. Green (a pediatrician) wanted to test if there is a correlation between the number of meals consumed by a child per day (X) and the child weight (Y). Included you will find a table containing the information on 6 of the children. Use the table to answer the following:
Child 
Number of meals consumed per day (X) 
child weight (Y) 
X² 
Y² 
XY 
Ahmad 
11 
8 
121 
64 
88 
Ali 
16 
11 
256 
121 
176 
Osama 
12 
9 
144 
81 
108 
Husien 
19 
13 
361 
169 
247 
Lala 
6 
8 
36 
64 
48 
Noor 





1. Find the average and standard deviation for the child number of meals and weight.
Answer:
Average no of meals= (11+16+12+19+6+0)/6 =64/6= 10.67
Average weight = (8+11+9+13+8+0)/6 = 49/6 = 8.17
Standard deviation of meals= square root of 1/6*236 = 6.27
Standard deviation of weights = square root of 1/6*99 = 4.06
2. Dose Ahmad has a better position among the 6 children in number of meals he consumed or in his weight? (Hint: use Z score formula to make the comparisons), justify your answer.
Answer;
Z score = (x –mean)/standard deviation
Meals = (1110.6)/6.27 = 0.638
Weights = (88.16)/4.06 = 0.039 = 0.04
Therefore, Ahmad has a better position in number of meals than in weights because the z score of meals is positive while that of weights is negative.
3. If these six children represent a sample from a population of 100 children what will be the number of children who will have a number of meals between 10 and 14.
Z score = (1010.6)/6.27 =0.0956 =0.1
Therefore (1410.6)/6.27 = 0.54
0.1*100 =10
0.54*100 = 54, therefore 54—10=64 children will have meals between 10 and 14
4. Is there a correlation between the two variables? Find the value of correlation coefficient.
Answer;
Yes there is correlation
R= sum (xmean of x) (ymean of y)/square root (xmean of x) ^2 *square root(ymean of y) ^2
=144/square root 236 * square root of 99
=144/15.36*9.95
=144/152.83 =0.94
R=0.94
5. Find out the prediction equation then use it to calculate the exact expected weight of a child that consumed 10 meals per day.
Answer;
8.16 weight = 10.6 meals
Therefore, 10 meals =?
= (8.16*10)/10.6 =81.6/10.6
=7.69
Approximate weight is 8
Q.3. A hospital supervisor has found that the medical stuff of the laboratory, on the average, complete a certain task in 10 minutes. If the times required to complete the task are approximately normally distributed with a standard deviation of 3 minutes, find:
1. The proportion of workers that will require more than 8 minutes to complete the task.
Answer:
(810)/3= 2/3 = 0.67 therefore 67%
2. If the supervisor has a total of 160 workers in his lab and any worker that completes the work in more than 15 minutes will be dismissed what is the total number of workers to be dismissed.
Answer:
Average time = 10, standard deviation 3
Expected time for all workers to finish with a standard deviation of 3
=10+3 minutes, 10+3=13, therefore no employees are expected to finish past 15 minutes
3. If a new machine was inserted which reduces the average time required from the original average by 2 minutes whereas the standard deviation is reduced by 1 minute, what will be the reduction in the proportion of workers that takes 6 minutes to finish the work. (Hint: find the proportion based on the old average and standard deviation then based on the new ones and then complete your answer).
Answer:
New average=8, new standard deviation = 2 minutes
(68)/2= 1
Q.4. A population of 1000 children have been surveyed for their hemoglobin level. The population average hemoglobin was 12 and their standard deviation was 3. Based on this information find out the following:
1. If a child that has a hemoglobin level below 11 will need iron supplementation what proportion will need iron?
Z11=(x12)/3=11, 33=x12, 33+12=x =45
Proportion = 45/1000= 0.045
2. What propotion of children will have a hemoglobin between 15.5 and 11?
Z 15.5= (x12)/3, =46.5=x12, =46.5+12=58.5=59
Proportion = 59/10000.045=0.0590.045 =0.014
3. What is the Z score for a child that has a hemoglobin level of 13
Z score = (1312)/3=1/3=0.33
4. What is the number of children that will have a hemoglobin level above 13?
Answer;
= (10.33)*1000= 670 children
5. What proportion of children will have a Z score below 2.5?
Answer;
2.5= (x12)/3=7.5+12=x, = 4.5=5children
Q.5. Answer the following:
 The least square prediction line is called so because it minimizes the total predictive errors. Answer by true or false______true______
 Presence of a strong correlation between two variables means that the pairs of the variables tend to occupy similar relative positions in their respective distributions. Answer by true or false__________true________
 The standard normal curve has an average of zero (0) and a standard deviation of 1.96. Answer by true or false______true______
 When correlation is found between two variables this mean that there is a causative relationship between them. Answer by true or false and justify your answer.
Answer;
True because for there to be a correlation between two variables a relationship must exist which can be either positive or negative.
 The graph that is used to represent quantitative data and at the same time keeps the individual observations is the frequency polygon. True or False? true
 Bar charts are not suitable to describe continuous quantitative variables, why?
answer;
Bar charts cannot illustrate a trend in continuous quantitative variables but rather reflect final definitive data.
Q.6 For each situation described by the statements below, what is the variable being measured? and what type of scale is used for its measurement? (Nominal, ordinal, interval or ratio).
Statement 
Variable measured 
Scale of measurement

1. In order to study the effect of different types of treatments on the recovery status the researcher classified the patients recovery status as quick, moderate, slow and late. 
time 
interval 
2. In a survey for health management purposes a researcher measured the time required by doctors to finish certain surgical operations. 
time 
ordinal 
3. In a study on type of disease the researcher has to classify the disease as genetic or not genetic 
disease 
nominal 
4. In a study of the effect of certain drugs of temperature the mean temperature of children in each treatment group were compared with the others. 
temperature 
ratio 
Statistics Problem Solving. Custom Statistics Problem Solving Essay Writing Service  Statistics Problem Solving Essay samples, help
Related essays
 Establish Criteria for Possible Solutions
 Should College Athletes be Exempted from Taking Exams
 Cyberbullying in the School System
 The Practice of doing Mathematics