Activity Data Type 1.Number of beatings from Wife Discrete 2.Results of rolling a dice Discrete 3.Weight of a person Continuous 4.Weight of Gold Continuous 5.Distance between two places Continuous 6.Length of a leaf Continuous 7.Dog's weight Continuous 8.Blue Color Discrete 9.Number of kids Discrete 10.Number of tickets in Indian railways Discrete 11.Number of times married Discrete 12.Gender (Male or Female) Discrete
Data Data Type 1.Gender Nominal 2.High School Class Ranking Ordinal 3.Celsius Temperature Interval 4.Weight Ratio 5.Hair Color Nominal 6.Socioeconomic Status Ordinal 7.Fahrenheit Temperature Interval 8.Height Ratio 9.Type of living accommodation Ordinal 10.Level of Agreement Ordinal 11.IQ(Intelligence Scale) Ratio 12.Sales Figures Ratio 13.Blood Group Nominal 15.Time Of Day Ordinal 16.Time on a Clock with Hands Interval 17.Number of Children Nominal 18.Religious Preference Nominal 19.Barometer Pressure Interval 20.SAT Scores Interval 21.Years of Education Ordinal
ANS- Three coins are tossed, Total number of Sample possible combinations = 23 = 8 Sample= {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} Number of combinations that have two heads and one tail = 3, (HHT, HTH, TTH) The probability of two heads and one tail when three coins are tossed simultaneously are P (Two heads and One tail) = Two heads and One tail /Number of desired outcomes = ⅜ or 0.375
a) Equal to 1 b) Less than or equal to 4 c) Sum is divisible by 2 and 3 ANS- for Two Dice total number of outcome -36 a) Equal to is 0 , As minimum sum is 2 for outcome (1,1) Hence Probability is 0 b) Less than or equal to 4: The possible combinations for sums less than or equal to 4 are: (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1). Probability = (Less than or equal to 4) / (Total Number of Outcomes) = 6 / 36 = 1 / 6 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 = (2 + 3 + 2) = 7 Let S be the sample space. Then, n(S) = Number of drawing 2 balls out of 7 =7C2 = (2×1)(7×6) =21 Let E = drawing 2 balls, none of which is blue. ∴n(E)= Number of drawing 2 balls out of (2 + 3) balls. =5C2 = (2×1)(5×4) =10 P(E)= n(E)/n(S) = 10/21 =0.476
Below are the probabilities of count of candies for children (ignoring the nature of the child-Generalized view)
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 = Expected Value = Σ (Value * Probability) Expected Value = (1 * 0.015) + (4 * 0.20) + (3 * 0.65) + (5 * 0.005) + (6 * 0.01) + (2 * 0.120) = 0.015 + 0.80 + 1.95 + 0.025 + 0.06 + 0.24 = 3.10
Q7) Calculate Mean, Median, Mode, Variance, Standard Deviation, Range & comment about the values / draw inferences, for the given dataset
###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]
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: weights of patients =108, 110, 123, 134, 135, 145, 167, 187, 199 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
Cars speed and distance Use Q9_a.csv
SP and Weight(WT) Use Q9_b.csv
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?
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]
Mean =41, Median =40.5, Variance =25.52 and Standard Deviation =5.05
Don’t have outliers and the data is slightly skewed towards right because mean is greater than median
The average score is around 41.0.
The middle value is 40.5, which is close to the mean.
Variance and Standard Deviation: The variance and standard deviation are high
ANS : When the mean and median of a dataset are equal, it indicates that the data has a symmetric distribution. The mean is at the center of the distribution. The median is also at the center of the distribution. The values on one side of the center are similar to the values on the other side.
ANS : When the mean is greater than the median, the distribution of the data is right-skewed or positively skewed.
ANS : When the median is greater than the mean, the distribution is said to be negatively skewed. This means that the tail of the distribution is stretched out to the left.
ANS : Data with positive kurtosis have more outliers than a normal distribution.
ANS : Negative kurtosis value indicates that wider peak and thinner tails.
What can we say about the distribution of the data?
ANS -Not normally distributed
What is nature of skewness of the data?
ANS- -Negative skewness
What will be the IQR of the data (approximately)?
ANS- -10-18
ANS. Both Boxplot 1 and Boxplot 2 have a similar interquartile range Boxplot 1 have a narrow range between the minimum and maximum values compared to Boxplot 2. & middle values of the data are similar in both plots
Data _set: Cars.csv Calculate the probability of MPG of Cars for the below cases. MPG <- Cars$MPG a. P(MPG>38) b. P(MPG<40) c. P (20<MPG<50) d. ANS. P(MPG>38)
Ans: a. P(MPG>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)= 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 (20<MPG<50)
P (20<MPG )= np.round(1-stats.norm.cdf(20, loc = q20.MPG.mean(), scale = q20.MPG.std()),3)
print('p(MPG>20)=',(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
P (20<MPG<50)= (prob_MPG_less_than_50) - (prob_MPG_greater_than_20)
print('P(20<MPG<50)=',(prob_MPG_greaterthan20_and_lessthan50))
P(20<MPG<50)= 0.013000000000000012
Dataset: Cars.csv
b) Check Whether the Adipose Tissue (AT) and Waist Circumference(Waist) from wc-at data set follows Normal Distribution
Dataset: wc-at.csv
Q 22) Calculate the Z scores of 90% confidence interval,94% confidence interval, 60% confidence interval
Q 23) Calculate the t scores of 95% confidence interval, 96% confidence interval, 99% confidence interval for sample size of 25
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
ANS
rcode = pt(tscore,df)
df degrees of freedom
Ans: import numpy as np
Import scipy as stats
t_score = (x - pop mean) / (sample standard daviation / square root of sample size)
(260-270)/90/np.sqrt(18))
_score = -0.471
stats.t.cdf(t_score, df = 17)
0.32 = 32%