histogram of the given data and also plot the relative frequencies and explain the distribution of weights of newborn babies.

Question

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Answer 1

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 8, Problem 8.1CRE

a.

To determine

To find: histogram of the given data and also plot the relative frequencies and explain the distribution of weights of newborn babies.

a.

Expert Solution

Explanation of Solution

Given:

The dataset:

Weight(Grams)	Count
Less than 500	5980
500 - 999	22015
1000 - 1499	29846
1500 - 1999	63427
2000 - 2499	204295
2500 - 2999	744181
3000 - 3499	1566755
3500 - 3999	1055004
4000 - 4499	262997
4500 - 4999	36706
5000 - 5499	4216

Formula used:

Relative Frequency = Frequency of classTotal frequency

Calculation:

Weight(Grams)	Relative frequency
Less than 500	0.001
500 - 999	0.006
1000 - 1499	0.007
1500 - 1999	0.016
2000 - 2499	0.051
2500 - 2999	0.186
3000 - 3499	0.392
3500 - 3999	0.264
4000 - 4499	0.066
4500 - 4999	0.009
5000 - 5499	0.001
Sum	1

Graph:

The histogram:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 8, Problem 8.1CRE , additional homework tip 1

The relative frequency histogram:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 8, Problem 8.1CRE , additional homework tip 2

Based on the above histogram, it looks that data is approximately normally distributed, though it has a bit larger tail towards the left side, so it can also be considered as the left skewed but since there is not so large difference in tails so it will be better to consider it as approximately normally distributed.

b.

To determine

To find the median class of the data.

b.

Expert Solution

Explanation of Solution

Formula used:

Median = (N2)Th term

Calculation:

Weight(Grams)	Count	Cumulative frequency.
Less than 500	5980	5980
500 - 999	22015	27995
1000 - 1499	29846	57841
1500 - 1999	63427	121268
2000 - 2499	204295	325563
2500 - 2999	744181	1069744
3000 - 3499	1566755	2636499
3500 - 3999	1055004	3691503
4000 - 4499	262997	3954500
4500 - 4999	36706	3991206
5000 - 5499	4216	3995422
Sum	3995422

Median = N2

Median = 39954222=1997711

Now, it required to see that in under which CF value does it falls.

The value 1997711 is less than 2636499, so the median value falls in the class 3000 − 3499.

c.

To determine

To construct the pie chart or bar chart to show the proportion of given categories.

c.

Expert Solution

Explanation of Solution

Given:

The categories:

Very low birth weight: Less than 1500 grams

Low birth weight: Less than 2500 grams

Normal birth weight: More than 2500 grams

Calculation:

Count of each category:

Category	Frequency	Proportion
Very low birth Weight	57841	0.0144768
Low birth weight	267722	0.0670072
Normal birth weight	3669859	0.918516

Graph:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 8, Problem 8.1CRE , additional homework tip 3

d.

To determine

To explain the outcome of the analysis that is being done so far.

d.

Expert Solution

Explanation of Solution

Based on the graph prepared in the last part, it is clearly visible that about 8% child having risk of low weight at the time of birth, which indicates at huge proportion of total, it is not an unusual events of being a low weight baby at the time of birth. Tough the distribution if approximately normal but still a large proportion is under the risk of low weight.

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Students have asked these similar questions

A population that is uniformly distributed between a=0and b=10 is given in sample sizes 50( ), 100( ), 250( ), and 500( ). Find the sample mean and the sample standard deviations for the given data. Compare your results to the average of means for a sample of size 10, and use the empirical rules to analyze the sampling error. For each sample, also find the standard error of the mean using formula given below. Standard Error of the Mean =sigma/Root Complete the following table with the results from the sampling experiment. (Round to four decimal places as needed.) Sample Size Average of 8 Sample Means Standard Deviation of 8 Sample Means Standard Error 50 100 250 500

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I need help with this problem and an explanation of the solution for the image described below. (Statistics: Engineering Probabilities)

Answer 2

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 8, Problem 8.1CRE

a.

To determine

To find: histogram of the given data and also plot the relative frequencies and explain the distribution of weights of newborn babies.

a.

Expert Solution

Explanation of Solution

Given:

The dataset:

Weight(Grams)	Count
Less than 500	5980
500 - 999	22015
1000 - 1499	29846
1500 - 1999	63427
2000 - 2499	204295
2500 - 2999	744181
3000 - 3499	1566755
3500 - 3999	1055004
4000 - 4499	262997
4500 - 4999	36706
5000 - 5499	4216

Formula used:

Relative Frequency = Frequency of classTotal frequency

Calculation:

Weight(Grams)	Relative frequency
Less than 500	0.001
500 - 999	0.006
1000 - 1499	0.007
1500 - 1999	0.016
2000 - 2499	0.051
2500 - 2999	0.186
3000 - 3499	0.392
3500 - 3999	0.264
4000 - 4499	0.066
4500 - 4999	0.009
5000 - 5499	0.001
Sum	1

Graph:

The histogram:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 8, Problem 8.1CRE , additional homework tip 1

The relative frequency histogram:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 8, Problem 8.1CRE , additional homework tip 2

Based on the above histogram, it looks that data is approximately normally distributed, though it has a bit larger tail towards the left side, so it can also be considered as the left skewed but since there is not so large difference in tails so it will be better to consider it as approximately normally distributed.

b.

To determine

To find the median class of the data.

b.

Expert Solution

Explanation of Solution

Formula used:

Median = (N2)Th term

Calculation:

Weight(Grams)	Count	Cumulative frequency.
Less than 500	5980	5980
500 - 999	22015	27995
1000 - 1499	29846	57841
1500 - 1999	63427	121268
2000 - 2499	204295	325563
2500 - 2999	744181	1069744
3000 - 3499	1566755	2636499
3500 - 3999	1055004	3691503
4000 - 4499	262997	3954500
4500 - 4999	36706	3991206
5000 - 5499	4216	3995422
Sum	3995422

Median = N2

Median = 39954222=1997711

Now, it required to see that in under which CF value does it falls.

The value 1997711 is less than 2636499, so the median value falls in the class 3000 − 3499.

c.

To determine

To construct the pie chart or bar chart to show the proportion of given categories.

c.

Expert Solution

Explanation of Solution

Given:

The categories:

Very low birth weight: Less than 1500 grams

Low birth weight: Less than 2500 grams

Normal birth weight: More than 2500 grams

Calculation:

Count of each category:

Category	Frequency	Proportion
Very low birth Weight	57841	0.0144768
Low birth weight	267722	0.0670072
Normal birth weight	3669859	0.918516

Graph:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 8, Problem 8.1CRE , additional homework tip 3

d.

To determine

To explain the outcome of the analysis that is being done so far.

d.

Expert Solution

Explanation of Solution

Based on the graph prepared in the last part, it is clearly visible that about 8% child having risk of low weight at the time of birth, which indicates at huge proportion of total, it is not an unusual events of being a low weight baby at the time of birth. Tough the distribution if approximately normal but still a large proportion is under the risk of low weight.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 8, Problem 8.1CRE

a.

To determine

To find: histogram of the given data and also plot the relative frequencies and explain the distribution of weights of newborn babies.

a.

Expert Solution

Explanation of Solution

Given:

The dataset:

Weight(Grams)	Count
Less than 500	5980
500 - 999	22015
1000 - 1499	29846
1500 - 1999	63427
2000 - 2499	204295
2500 - 2999	744181
3000 - 3499	1566755
3500 - 3999	1055004
4000 - 4499	262997
4500 - 4999	36706
5000 - 5499	4216

Formula used:

Relative Frequency = Frequency of classTotal frequency

Calculation:

Weight(Grams)	Relative frequency
Less than 500	0.001
500 - 999	0.006
1000 - 1499	0.007
1500 - 1999	0.016
2000 - 2499	0.051
2500 - 2999	0.186
3000 - 3499	0.392
3500 - 3999	0.264
4000 - 4499	0.066
4500 - 4999	0.009
5000 - 5499	0.001
Sum	1

Graph:

The histogram:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 8, Problem 8.1CRE , additional homework tip 1

The relative frequency histogram:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 8, Problem 8.1CRE , additional homework tip 2

Based on the above histogram, it looks that data is approximately normally distributed, though it has a bit larger tail towards the left side, so it can also be considered as the left skewed but since there is not so large difference in tails so it will be better to consider it as approximately normally distributed.

b.

To determine

To find the median class of the data.

b.

Expert Solution

Explanation of Solution

Formula used:

Median = (N2)Th term

Calculation:

Weight(Grams)	Count	Cumulative frequency.
Less than 500	5980	5980
500 - 999	22015	27995
1000 - 1499	29846	57841
1500 - 1999	63427	121268
2000 - 2499	204295	325563
2500 - 2999	744181	1069744
3000 - 3499	1566755	2636499
3500 - 3999	1055004	3691503
4000 - 4499	262997	3954500
4500 - 4999	36706	3991206
5000 - 5499	4216	3995422
Sum	3995422

Median = N2

Median = 39954222=1997711

Now, it required to see that in under which CF value does it falls.

The value 1997711 is less than 2636499, so the median value falls in the class 3000 − 3499.

c.

To determine

To construct the pie chart or bar chart to show the proportion of given categories.

c.

Expert Solution

Explanation of Solution

Given:

The categories:

Very low birth weight: Less than 1500 grams

Low birth weight: Less than 2500 grams

Normal birth weight: More than 2500 grams

Calculation:

Count of each category:

Category	Frequency	Proportion
Very low birth Weight	57841	0.0144768
Low birth weight	267722	0.0670072
Normal birth weight	3669859	0.918516

Graph:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 8, Problem 8.1CRE , additional homework tip 3

d.

To determine

To explain the outcome of the analysis that is being done so far.

d.

Expert Solution

Explanation of Solution

Based on the graph prepared in the last part, it is clearly visible that about 8% child having risk of low weight at the time of birth, which indicates at huge proportion of total, it is not an unusual events of being a low weight baby at the time of birth. Tough the distribution if approximately normal but still a large proportion is under the risk of low weight.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 8, Problem 8.1CRE

a.

To determine

To find: histogram of the given data and also plot the relative frequencies and explain the distribution of weights of newborn babies.

a.

Expert Solution

Explanation of Solution

Given:

The dataset:

Weight(Grams)	Count
Less than 500	5980
500 - 999	22015
1000 - 1499	29846
1500 - 1999	63427
2000 - 2499	204295
2500 - 2999	744181
3000 - 3499	1566755
3500 - 3999	1055004
4000 - 4499	262997
4500 - 4999	36706
5000 - 5499	4216

Formula used:

Relative Frequency = Frequency of classTotal frequency

Calculation:

Weight(Grams)	Relative frequency
Less than 500	0.001
500 - 999	0.006
1000 - 1499	0.007
1500 - 1999	0.016
2000 - 2499	0.051
2500 - 2999	0.186
3000 - 3499	0.392
3500 - 3999	0.264
4000 - 4499	0.066
4500 - 4999	0.009
5000 - 5499	0.001
Sum	1

Graph:

The histogram:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 8, Problem 8.1CRE , additional homework tip 1

The relative frequency histogram:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 8, Problem 8.1CRE , additional homework tip 2

Based on the above histogram, it looks that data is approximately normally distributed, though it has a bit larger tail towards the left side, so it can also be considered as the left skewed but since there is not so large difference in tails so it will be better to consider it as approximately normally distributed.

b.

To determine

To find the median class of the data.

b.

Expert Solution

Explanation of Solution

Formula used:

Median = (N2)Th term

Calculation:

Weight(Grams)	Count	Cumulative frequency.
Less than 500	5980	5980
500 - 999	22015	27995
1000 - 1499	29846	57841
1500 - 1999	63427	121268
2000 - 2499	204295	325563
2500 - 2999	744181	1069744
3000 - 3499	1566755	2636499
3500 - 3999	1055004	3691503
4000 - 4499	262997	3954500
4500 - 4999	36706	3991206
5000 - 5499	4216	3995422
Sum	3995422

Median = N2

Median = 39954222=1997711

Now, it required to see that in under which CF value does it falls.

The value 1997711 is less than 2636499, so the median value falls in the class 3000 − 3499.

c.

To determine

To construct the pie chart or bar chart to show the proportion of given categories.

c.

Expert Solution

Explanation of Solution

Given:

The categories:

Very low birth weight: Less than 1500 grams

Low birth weight: Less than 2500 grams

Normal birth weight: More than 2500 grams

Calculation:

Count of each category:

Category	Frequency	Proportion
Very low birth Weight	57841	0.0144768
Low birth weight	267722	0.0670072
Normal birth weight	3669859	0.918516

Graph:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 8, Problem 8.1CRE , additional homework tip 3

d.

To determine

To explain the outcome of the analysis that is being done so far.

d.

Expert Solution

Explanation of Solution

Based on the graph prepared in the last part, it is clearly visible that about 8% child having risk of low weight at the time of birth, which indicates at huge proportion of total, it is not an unusual events of being a low weight baby at the time of birth. Tough the distribution if approximately normal but still a large proportion is under the risk of low weight.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 8, Problem 8.1CRE

a.

To determine

To find: histogram of the given data and also plot the relative frequencies and explain the distribution of weights of newborn babies.

a.

Expert Solution

Explanation of Solution

Given:

The dataset:

Weight(Grams)	Count
Less than 500	5980
500 - 999	22015
1000 - 1499	29846
1500 - 1999	63427
2000 - 2499	204295
2500 - 2999	744181
3000 - 3499	1566755
3500 - 3999	1055004
4000 - 4499	262997
4500 - 4999	36706
5000 - 5499	4216

Formula used:

Relative Frequency = Frequency of classTotal frequency

Calculation:

Weight(Grams)	Relative frequency
Less than 500	0.001
500 - 999	0.006
1000 - 1499	0.007
1500 - 1999	0.016
2000 - 2499	0.051
2500 - 2999	0.186
3000 - 3499	0.392
3500 - 3999	0.264
4000 - 4499	0.066
4500 - 4999	0.009
5000 - 5499	0.001
Sum	1

Graph:

The histogram:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 8, Problem 8.1CRE , additional homework tip 1

The relative frequency histogram:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 8, Problem 8.1CRE , additional homework tip 2

Based on the above histogram, it looks that data is approximately normally distributed, though it has a bit larger tail towards the left side, so it can also be considered as the left skewed but since there is not so large difference in tails so it will be better to consider it as approximately normally distributed.

b.

To determine

To find the median class of the data.

b.

Expert Solution

Explanation of Solution

Formula used:

Median = (N2)Th term

Calculation:

Weight(Grams)	Count	Cumulative frequency.
Less than 500	5980	5980
500 - 999	22015	27995
1000 - 1499	29846	57841
1500 - 1999	63427	121268
2000 - 2499	204295	325563
2500 - 2999	744181	1069744
3000 - 3499	1566755	2636499
3500 - 3999	1055004	3691503
4000 - 4499	262997	3954500
4500 - 4999	36706	3991206
5000 - 5499	4216	3995422
Sum	3995422

Median = N2

Median = 39954222=1997711

Now, it required to see that in under which CF value does it falls.

The value 1997711 is less than 2636499, so the median value falls in the class 3000 − 3499.

c.

To determine

To construct the pie chart or bar chart to show the proportion of given categories.

c.

Expert Solution

Explanation of Solution

Given:

The categories:

Very low birth weight: Less than 1500 grams

Low birth weight: Less than 2500 grams

Normal birth weight: More than 2500 grams

Calculation:

Count of each category:

Category	Frequency	Proportion
Very low birth Weight	57841	0.0144768
Low birth weight	267722	0.0670072
Normal birth weight	3669859	0.918516

Graph:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 8, Problem 8.1CRE , additional homework tip 3

d.

To determine

To explain the outcome of the analysis that is being done so far.

d.

Expert Solution

Explanation of Solution

Based on the graph prepared in the last part, it is clearly visible that about 8% child having risk of low weight at the time of birth, which indicates at huge proportion of total, it is not an unusual events of being a low weight baby at the time of birth. Tough the distribution if approximately normal but still a large proportion is under the risk of low weight.

Answer 6

Chapter 8, Problem 8.1CRE

a.

To determine

To find: histogram of the given data and also plot the relative frequencies and explain the distribution of weights of newborn babies.

a.

Expert Solution

Explanation of Solution

Given:

The dataset:

Weight(Grams)	Count
Less than 500	5980
500 - 999	22015
1000 - 1499	29846
1500 - 1999	63427
2000 - 2499	204295
2500 - 2999	744181
3000 - 3499	1566755
3500 - 3999	1055004
4000 - 4499	262997
4500 - 4999	36706
5000 - 5499	4216

Formula used:

Relative Frequency = Frequency of classTotal frequency

Calculation:

Weight(Grams)	Relative frequency
Less than 500	0.001
500 - 999	0.006
1000 - 1499	0.007
1500 - 1999	0.016
2000 - 2499	0.051
2500 - 2999	0.186
3000 - 3499	0.392
3500 - 3999	0.264
4000 - 4499	0.066
4500 - 4999	0.009
5000 - 5499	0.001
Sum	1

Graph:

The histogram:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 8, Problem 8.1CRE , additional homework tip 1

The relative frequency histogram:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 8, Problem 8.1CRE , additional homework tip 2

Based on the above histogram, it looks that data is approximately normally distributed, though it has a bit larger tail towards the left side, so it can also be considered as the left skewed but since there is not so large difference in tails so it will be better to consider it as approximately normally distributed.

Answer 7

Chapter 8, Problem 8.1CRE

a.

To determine

To find: histogram of the given data and also plot the relative frequencies and explain the distribution of weights of newborn babies.

Answer 8

a.

Expert Solution

Explanation of Solution

Given:

The dataset:

Weight(Grams)	Count
Less than 500	5980
500 - 999	22015
1000 - 1499	29846
1500 - 1999	63427
2000 - 2499	204295
2500 - 2999	744181
3000 - 3499	1566755
3500 - 3999	1055004
4000 - 4499	262997
4500 - 4999	36706
5000 - 5499	4216

Formula used:

Relative Frequency = Frequency of classTotal frequency

Calculation:

Weight(Grams)	Relative frequency
Less than 500	0.001
500 - 999	0.006
1000 - 1499	0.007
1500 - 1999	0.016
2000 - 2499	0.051
2500 - 2999	0.186
3000 - 3499	0.392
3500 - 3999	0.264
4000 - 4499	0.066
4500 - 4999	0.009
5000 - 5499	0.001
Sum	1

Graph:

The histogram:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 8, Problem 8.1CRE , additional homework tip 1

The relative frequency histogram:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 8, Problem 8.1CRE , additional homework tip 2

Based on the above histogram, it looks that data is approximately normally distributed, though it has a bit larger tail towards the left side, so it can also be considered as the left skewed but since there is not so large difference in tails so it will be better to consider it as approximately normally distributed.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

b.

To determine

To find the median class of the data.

b.

Expert Solution

Explanation of Solution

Formula used:

Median = (N2)Th term

Calculation:

Weight(Grams)	Count	Cumulative frequency.
Less than 500	5980	5980
500 - 999	22015	27995
1000 - 1499	29846	57841
1500 - 1999	63427	121268
2000 - 2499	204295	325563
2500 - 2999	744181	1069744
3000 - 3499	1566755	2636499
3500 - 3999	1055004	3691503
4000 - 4499	262997	3954500
4500 - 4999	36706	3991206
5000 - 5499	4216	3995422
Sum	3995422

Median = N2

Median = 39954222=1997711

Now, it required to see that in under which CF value does it falls.

The value 1997711 is less than 2636499, so the median value falls in the class 3000 − 3499.

Answer 13

b.

To determine

To find the median class of the data.

Answer 14

b.

Expert Solution

Explanation of Solution

Formula used:

Median = (N2)Th term

Calculation:

Weight(Grams)	Count	Cumulative frequency.
Less than 500	5980	5980
500 - 999	22015	27995
1000 - 1499	29846	57841
1500 - 1999	63427	121268
2000 - 2499	204295	325563
2500 - 2999	744181	1069744
3000 - 3499	1566755	2636499
3500 - 3999	1055004	3691503
4000 - 4499	262997	3954500
4500 - 4999	36706	3991206
5000 - 5499	4216	3995422
Sum	3995422

Median = N2

Median = 39954222=1997711

Now, it required to see that in under which CF value does it falls.

The value 1997711 is less than 2636499, so the median value falls in the class 3000 − 3499.

Answer 15

b.

Expert Solution

Answer 16

b.

Expert Solution

Answer 17

Expert Solution

Answer 18

c.

To determine

To construct the pie chart or bar chart to show the proportion of given categories.

c.

Expert Solution

Explanation of Solution

Given:

The categories:

Very low birth weight: Less than 1500 grams

Low birth weight: Less than 2500 grams

Normal birth weight: More than 2500 grams

Calculation:

Count of each category:

Category	Frequency	Proportion
Very low birth Weight	57841	0.0144768
Low birth weight	267722	0.0670072
Normal birth weight	3669859	0.918516

Graph:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 8, Problem 8.1CRE , additional homework tip 3

Answer 19

c.

To determine

To construct the pie chart or bar chart to show the proportion of given categories.

Answer 20

c.

Expert Solution

Explanation of Solution

Given:

The categories:

Very low birth weight: Less than 1500 grams

Low birth weight: Less than 2500 grams

Normal birth weight: More than 2500 grams

Calculation:

Count of each category:

Category	Frequency	Proportion
Very low birth Weight	57841	0.0144768
Low birth weight	267722	0.0670072
Normal birth weight	3669859	0.918516

Graph:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 8, Problem 8.1CRE , additional homework tip 3

Answer 21

c.

Expert Solution

Answer 22

c.

Expert Solution

Answer 23

Expert Solution

Answer 24

d.

To determine

To explain the outcome of the analysis that is being done so far.

d.

Expert Solution

Explanation of Solution

Based on the graph prepared in the last part, it is clearly visible that about 8% child having risk of low weight at the time of birth, which indicates at huge proportion of total, it is not an unusual events of being a low weight baby at the time of birth. Tough the distribution if approximately normal but still a large proportion is under the risk of low weight.

Answer 25

d.

To determine

To explain the outcome of the analysis that is being done so far.

Answer 26

d.

Expert Solution

Explanation of Solution

Based on the graph prepared in the last part, it is clearly visible that about 8% child having risk of low weight at the time of birth, which indicates at huge proportion of total, it is not an unusual events of being a low weight baby at the time of birth. Tough the distribution if approximately normal but still a large proportion is under the risk of low weight.