some of the statictis to test : Q1) Identify the Data type for the Following:

Number of beatings from Wife Discrete Results of rolling a dice Discrete Weight of a person Continuous Weight of Gold Continuous Distance between two places Continuous Length of a leaf Continuous Dog's weight Continuous Blue Color Discrete Number of kids Discrete Number of tickets in Indian railways Discrete Number of times married Discrete Gender (Male or Female) Discrete

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 scale Weight Ratio Hair Color Nominal Socioeconomic Status Ordinal Fahrenheit Temperature Interval scale Height Ratio Type of living accommodation Nominal Level of Agreement Ordinal IQ(Intelligence Scale) Ratio Sales Figures Ratio Blood Group Nominal Time Of Day Interval Time on a Clock with Hands Interval Number of Children Nominal Religious Preference Nominal Barometer Pressure Interval SAT Scores Ordinal Years of Education Interval

Q3) Three Coins are tossed, find the probability that two heads and one tail are obtained? Total no of possibilities= 23 = 8 Total no sample { (HHH), (HHT), (HTH), (THH), (HTT), (TTH), (THT), (TTT)} Probability of Two heads and one tail={(HTT), (TTH), (THT)} =3\8 The probability that two heads and one tail is 0.375 (or) 37.5%

Q4) Two Dice are rolled, find the probability that sum is The sample space = 62 = 36 Equal to 1 The probability that the sum is equal to 1 is 0 Because there is no combination that the dice provides sum 1

Less than or equal to 4

The total possibilities of sum being less than or equal to 4 =6 The probability is = 6/36 =1/6 The probability that the sum is less than or equal to 4 is 0.166 (or) 16.6%

Sum is divisible by 2 and  3

The total possible outcomes divisible by 2 and 3 = 24 The probability is possible outcome/total outcome= 6/36=1/6 The probability that the Sum is divisible by 2 and 3= 0.166 (or) 16.6%

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? Solution: The total no of balls=(2+3+2)=7 The probability of the first ball is not blue = 5/7(“blue ball \ total no of balls”) The probability of the second ball is not blue = 4/6(“blue ball \ total no of balls”) probability that none of the balls drawn is blue P = (5/7)(4/6)=20/42 = 10/21 =0.476 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 Solution: Expected value = ∑probabilitycandies = (10.015)+ (40.20)+ (30.65)+ (50.005)+ (60.01)+(20.120) = 3.09 The Expected number of candies for a randomly selected child is 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. Solution: RESULTS Points Score Weigh Mean 3.596563 3.21725 17.84875 Median 3.695 3.325 17.71 Mode 3.92,3.07 3.44 17.02,18.90 Variance 0.285881 0.957379 3.193166 Standard Deviation 0.534679 0.978457 1.786943 Range 2.17 3.911 8.4

Inferences:
The columns points and Weigh is multimodal For the Points and score Median> mean , hence the distribution is Right skewed distribution For Weigh Mean> Median, hence the distribution is Left skewed distribution The boxplot is plotted below by which we can view Inter-Quartile Region and outliers

Q8) Calculate Expected Value for the problem below 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? Expected value = ∑probabilityweights = (1/9)[108+110+123+134+135+145+167+187+199) =1308/9 =145.33

Q9) Calculate Skewness, Kurtosis & draw inferences on the following data Cars speed and distance Use Q9_a.csv
From the above graph we can infer that the speed data is left skewed and it is platykurtic kurtosis (-ve kurtosis) i.e. flatter distribution.

From the above graph we can infer that the Distance data is right skewed(+ve ) and it is  leptokurtic kurtosis (+ve kurtosis) i.e. more peakdness distribution.

SP and Weight(WT) From the below graph we can infer that the SP data is right skewed(+ve ) and it is leptokurtic kurtosis (+ve kurtosis) i.e. more peakdness distribution. There is outlier values in the upper extreme zone.

From the below  graph we can infer that the WT data is left skewed and it is  leptokurtic kurtosis (+ve kurtosis) i.e. more peakdness distribution.
This data have outliers in the upper and lower extreme zones.

Use Q9_b.csv

Q10) Draw inferences about the following boxplot & histogram

This histogram represents  Right skewed distribution and in this case Median>mean

From this box plot we could infer that more no of outlier values are present in the upper extreme zone.
This graph shows that it is right skewed
The range of whisker is very wide in the upper quartile region
1.5IQR gives the limit if the upper extreme point and the values of data beyond this upper extreme point are termed as Outliers .
These outliers are to be omitted in the distribution to obtain a normal distribution.

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? Solution: Interval= point estimate ± margin of error Point estimate =200 pounds Margin of error = (standard deviation)/(sqrt(samples)); Standard deviation is 200 ; The no of sample drawn is 2000

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 Find mean, median, variance, standard deviation. What can we say about the student marks?

From the students marks it is seen that the mean and median is almost equal and hence the data is symmetrically distributed   

This data have two outlier values which is beyond the upper quartile region(49,56)

Q13) What is the nature of skewness when mean, median of data are equal? In case of Mean=median , the nature of skewness is symmetrically distributed Q14) What is the nature of skewness when mean > median ? In case of Mean>median , the nature of skewness is Negative in nature which is left skewed distribution

Q15) What is the nature of skewness when median > mean? In case of Mean<median , the nature of skewness is Positive in nature which is Right skewed distribution

Q16) What does positive kurtosis value indicates for a data ? Positive kurtosis for a data Implies that the data is of more peaked distribution than the normal distribution Q17) What does negative kurtosis value indicates for a data? Negative kurtosis for a data Implies that the data is of flatter distribution than the normal distribution

Q18) Answer the below questions using the below boxplot visualization.

What can we say about the distribution of the data? The given data is not symmetrically distributed ,also more no of values lies below the lower quartile region What is nature of skewness of the data? This data is left skewed Distribution (-ve skewness) What will be the IQR of the data (approximately)? IQR = upper quartile-lower quartile =18-10 IQR = 8 Q19) Comment on the below Boxplot visualizations?

Draw an Inference from the distribution of data for Boxplot 1 with respect Boxplot 2. Answer: Box plot 2 has a more wider number of data than box plot 1 Boxplot 2 if flatter in distribution as compared to box plot 1 which is more peaked. Both these data are symmentrically distributed.

Q 20) 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 P(MPG>38) P(MPG<40) P (20<MPG<50)

Q 21) Check whether the data follows normal distribution Check whether the MPG of Cars follows Normal Distribution Dataset: Cars.csv Median >mean . hence it is left skewed distribution

b)Check Whether the Adipose Tissue (AT) and Waist Circumference(Waist) from wc-at data set follows Normal Distribution Dataset: wc-at.csv

Mean > Median , The distribution is Right Skewed distribution.

Q 22) Calculate the Z scores of 90% confidence interval,94% confidence interval, 60% confidence interval

Solution: For 90% confidence interval  A=(1+0.90)/2 = 0.95 For 94% confidence interval  A=(1+0.94)/2 = 0.97 For 60% confidence interval  A=(1+0.60)/2 = 0.80

        Q 23) Calculate the t scores of 95% confidence interval, 96% confidence interval, 99% confidence interval for sample size of 25

Solution: For 95% confidence interval  A=(1+0.95)/2 = 0.975 For 96% confidence interval  A=(1+0.96)/2 = 0.98 For 99% confidence interval  A=(1+0.99)/2 = 0.995

Q 24) A Government company claims that an average light bulb lasts 270 days. A researcher randomly selects 18 bulbs for testing. The sampled bulbs last an average of 260 days, with a standard deviation of 90 days. If the CEO's claim were true, what is the probability that 18 randomly selected bulbs would have an average life of no more than 260 days Hint:
rcode  pt(tscore,df)
df  degrees of freedom Solution: No of items in the sample(n)=18 ;df=n-1=17 Mean of the sampled bulbs(x)=260 Standard deviation(ơ)=90 Mean of population(µ)=270 t=(x-µ)/(s/√n)=(260-270)/(90/√18)= -0.471

The probability that 18 randomly selected bulbs would have an average life of no more than 260 days is 0.321