https://github.com/moindalvs/assignment-basic-stats-level1
Assignment Basic Stats
https://github.com/moindalvs/assignment-basic-stats-level1
bia binomial-distribution boxplot confidence-intervals datatypes descriptive-statistics inferential-statistics normal-distribution numpy pandas probability python3 scipy-stats statistics z-score
Last synced: 7 months ago
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Assignment Basic Stats
- Host: GitHub
- URL: https://github.com/moindalvs/assignment-basic-stats-level1
- Owner: MoinDalvs
- Created: 2022-02-15T13:26:54.000Z (almost 4 years ago)
- Default Branch: main
- Last Pushed: 2022-03-03T07:20:21.000Z (almost 4 years ago)
- Last Synced: 2025-02-17T09:34:34.014Z (10 months ago)
- Topics: bia, binomial-distribution, boxplot, confidence-intervals, datatypes, descriptive-statistics, inferential-statistics, normal-distribution, numpy, pandas, probability, python3, scipy-stats, statistics, z-score
- Language: Jupyter Notebook
- Homepage:
- Size: 439 KB
- Stars: 1
- Watchers: 1
- Forks: 3
- Open Issues: 0
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Metadata Files:
- Readme: README.md
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README
# Assignment Basic Stats
## Q1) Identify the Data type for the Following:
## Q2) Identify the Data types, which were among the following
### Nominal, Ordinal, Interval, Ratio.
Data Data Type
Gender Nominal
High School Class Ranking Ordinal
Celsius Temperature Interval
Weight Ratio
Hair Color Nominal
Socioeconomic Status Ordinal
Fahrenheit Temperature Interval
Height Ratio
Type of living accommodation Ordinal
Level of Agreement Ordinal
IQ (Intelligence Scale) Ratio
Sales Figures Ratio
Blood Group Nominal
Time Of Day Ordinal
Time on a Clock with Hands Interval
Number of Children Nominal
Religious Preference Nominal
Barometer Pressure Interval
SAT Scores Interval
Years of Education Ordinal
## Q3) Three Coins are tossed, find the probability that two heads and one tail are obtained?
Ans:
P (Two heads and one tail) = N (Event (Two heads and one tail)) / N (Event (Three
coins tossed))
= 3/8 = 0.375 = 37.5%
## Q4) Two Dice are rolled, find the probability where sum is
a) Equal to 1
b) Less than or equal to 4
c) Divisible by 2 and 3
Ans:
Number of possible outcomes for the above event is
N (Event (Two dice rolled)) = 6^2 = 36
a.) P (sum is Equal to 1) = ‘0’ zero null nada none.
b.) P (Sum is less than or equal to 4) = N (Event (Sum is less than or equal to 4)) / N (Event (Two dice rolled))
= 6 / 36 = 1/6 = 0.166 = 16.66%
c.) P (Sum is divisible by 2 and 3) = N (Event (Sum is divisible by 2 and 3)) / N(Event (Two dice rolled))
= 6 / 36 = 1/6 = 0.16 = 16.66%
## Q5) A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
Ans: Total number of balls =7 balls
N (Event (2 balls are drawn randomly from bag) = 7! / 2! * 5!
= (7*6*5*4*3*2*1) / (2*1) * (5*4*3*2*1)
N (Event (2 balls are drawn randomly from bag) = (7*6)/ (2*1) = 21
If none of them drawn 2 balls are blue = 7 – 2 = 5
N (Event (None of the balls drawn is blue) = 5! / 2! * 3! = (5*4) / (2*1)
= 10
P (None of the balls drawn is blue) = N (Event (None of the balls drawn is blue) / N (Event (2 balls are drawn randomly from bag)
= 10 / 21
## Q6) Calculate the Expected number of candies for a randomly selected child
Below are the probabilities of count of candies for children (ignoring the nature of the child-Generalized view)
CHILD Candies count Probability
A 1 0.015
B 4 0.20
C 3 0.65
D 5 0.005
E 6 0.01
F 2 0.120
Child A – probability of having 1 candy = 0.015.
Child B – probability of having 4 candies = 0.20
Ans:
0.015+0.8+1.95+0.025+0.06+0.24 = 3.09
## Q7) Calculate Mean, Median, Mode, Variance, Standard Deviation, Range & comment about the values / draw inferences, for the given dataset
- For Points, Score, Weigh>
Find Mean, Median, Mode, Variance, Standard Deviation, and Range and also Comment about the values/ Draw some inferences.
Use Q7.csv file
Ans:
Mean for Points = 3.59, Score = 3.21 and Weigh = 17.84
Median for Points = 3.69, Score = 3.32 and Weigh = 17.71
Mode for Points = 3.07, Score = 3.44 and Weigh = 17.02
Variance for Points = 0.28, Score = 0.95, Weigh = 3.19
Standard Deviation for Points = 0.53, Score = 0.97, Weigh = 1.78
Range [Min-Max] for Points [3.59 – 4.93], Score [3.21 – 5.42] and Weigh [17.84 – 22.9]
Draw Inferences
## Q8) Calculate Expected Value for the problem below
a) The weights (X) of patients at a clinic (in pounds), are
108, 110, 123, 134, 135, 145, 167, 187, 199
Assume one of the patients is chosen at random. What is the Expected Value of the Weight of that patient?
Ans: Expected value = Sum (X * Probability of X)
= (1/9)(108)+ (1/9)(110)+ (1/9)(123)+ (1/9)(134)+ (1/9)(145)+ (1/9)(167)+ (1/9)(187)+ (1/9)(199)
= 145.33
## Q9) Calculate Skewness, Kurtosis & draw inferences on the following data
Cars speed and distance
Use Q9_a.csv
Ans:
q9a = pd.read_csv("C:/Users/Moin Dalvi/Documents/EXcelR Study and Assignment Material/Data Science Assignments/Basic Statistics 1/Q9_a.csv", index_col = 'Index')
print('For Cars Speed', "Skewness value=", np.round(q9a.speed.skew(),2), 'and' , 'Kurtosis value=', np.round(q9a.dist.skew(),2))
For Cars Speed Skewness value= -0.12 and Kurtosis value= 0.81
print('Skewness value =', np.round(q9a.dist.skew(),2),'and', 'Kurtosis value =', np.round(q9a.dist.kurt(),2), 'for Cars Distance')
Skewness value = 0.81 and Kurtosis value = 0.41 for Cars Distance
SP and Weight (WT)
Use Q9_b.csv
Ans:
q9b =pd.read_csv("C:/Users/Moin Dalvi/Documents/EXcelR Study and Assignment Material/Data Science Assignments/Basic Statistics 1/Q9_b.csv")
print('For SP Skewness =', np.round(q9b.SP.skew(),2), 'kurtosis =', np.round(q9b.WT.kurt(),2))
For SP Skewness = 1.61 kurtosis = 0.95
print('For WT Skewness =', np.round(q9b.SP.skew(),2), 'Kurtosis =', np.round(q9b.WT.kurt(),2))
For WT Skewness = 1.61 Kurtosis = 0.95
## Q10) Draw inferences about the following boxplot & histogram
Ans: The histograms peak has right skew and tail is on right. Mean > Median. We have outliers on the higher side.
Ans: The boxplot has outliers on the maximum side.
## Q11) Suppose we want to estimate the average weight of an adult male in Mexico. We draw a random sample of 2,000 men from a population of 3,000,000 men and weigh them. We find that the average person in our sample weighs 200 pounds, and the standard deviation of the sample is 30 pounds. Calculate 94%,98%,96% confidence interval?
Ans:
conf_94 =stats.t.interval(alpha = 0.94, df=1999, loc=200, scale=30/np.sqrt(2000))
print(np.round(conf_94,0))
print(conf_94)
For 94% confidence interval Range is [ 198.73 – 201.26]
For 98% confidence interval range is [198.43 – 201.56]
For 96% confidence interval range is [198.62 – 201.37]
## Q12) Below are the scores obtained by a student in tests
34,36,36,38,38,39,39,40,40,41,41,41,41,42,42,45,49,56
1) Find mean, median, variance, standard deviation.
Ans: Mean =41, Median =40.5, Variance =25.52 and Standard Deviation =5.05
2) What can we say about the student marks?
Ans: we don’t have outliers and the data is slightly skewed towards right because mean is greater than median.
## Q13) What is the nature of skewness when mean and median of data are equal?
Ans: No skewness is present we have a perfect symmetrical distribution
## Q14) What is the nature of skewness when mean > median?
Ans: Skewness and tail is towards Right
## Q15) What is the nature of skewness when median > mean?
Ans: Skewness and tail is towards left
## Q16) What does positive kurtosis value indicates for a data?
Ans: Positive kurtosis means the curve is more peaked and it is Leptokurtic
## Q17) What does negative kurtosis value indicates for a data?
Ans: Negative Kurtosis means the curve will be flatter and broader
## Q18) Answer the below questions using the below boxplot visualization.
+ What can we say about the distribution of the data?
Ans: The above Boxplot is not normally distributed the median is towards the higher value
+ What is nature of skewness of the data?
Ans: The data is a skewed towards left. The whisker range of minimum value is greater than maximum
+ What will be the IQR of the data (approximately)?
Ans: The Inter Quantile Range = Q3 Upper quartile – Q1 Lower Quartile = 18 – 10 =8
## Q19) Comment on the below Boxplot visualizations?
Draw an Inference from the distribution of data for Boxplot 1 with respect Boxplot 2.
Ans: First there are no outliers. Second both the box plot shares the same median that is approximately in a range between 275 to 250 and they are normally distributed with zero to no skewness neither at the minimum or maximum whisker range.
## Q20) Calculate probability from the given dataset for the below cases
Data _set: Cars.csv
Calculate the probability of MPG of Cars for the below cases.
MPG <- Cars $ MPG
a. P(MPG>38)
Ans: Prob_MPG_greater_than_38 = np.round(1 - stats.norm.cdf(38, loc= q20.MPG.mean(), scale= q20.MPG.std()),3)
print('P(MPG>38)=',Prob_MPG_greater_than_38)
P(MPG>38)= 0.348
b. P(MPG<40)
Ans: prob_MPG_less_than_40 = np.round(stats.norm.cdf(40, loc = q20.MPG.mean(), scale = q20.MPG.std()),3)
print('P(MPG<40)=',prob_MPG_less_than_40)
P(MPG<40)= 0.729
c. P (2020)=',(prob_MPG_greater_than_20))
p(MPG>20)= 0.943
prob_MPG_less_than_50 = np.round(stats.norm.cdf(50, loc = q20.MPG.mean(), scale = q20.MPG.std()),3)
print('P(MPG<50)=',(prob_MPG_less_than_50))
P(MPG<50)= 0.956
prob_MPG_greaterthan20_and_lessthan50= (prob_MPG_less_than_50) - (prob_MPG_greater_than_20)
print('P(20