Math Gamification

Materials for foreign and Russian-speaking students studying the Data Analysis course in English

1. Независимо друг от друга 4 человека садятся в поезд, содержащий 13 вагонов. Найдите вероятность того, что все они поедут в разных вагонах

2. Внутрь круга радиуса 50 наудачу брошена точка. Найдите вероятность того, что точка окажется внутри вписанного в круг правильного шестиугольника.

3. События A, B и C независимы. Найдите вероятность того, что из событий A, B и C наступит ровно одно событие, если P(A)= 0,2, P(B)= 0,4 и P(C)= 0,9.

4. В первой урне m1=8 белых и n1=3 черных шаров, во второй – m2=7 белых и n2=8 черных. Из второй урны случайным образом перекладывают в первую два шара, после чего из первой урны берут один шар, который оказывается белым. Какова ероятность того, что два шара, переложенные из второй урны в первую, были разных цветов?

5. Вероятность выпуска бракованного изделия равна 0, 47. Не используя приближенную формулу для числа успехов в схеме Бернулли, найдите вероятность того, что среди 110 выпущенных изделий ровно 57 изделий без брака.

Test 1\ Контрольная работа № 1

Test 2 \ Контрольная работа № 2

1. Закон распределения дискретной случайной величины задан таблицей: X={-2,-1,2,6} p={0.2,0.1, ...., 0.3}

Найти математическое ожидание E(X) и дисперсию D(X). Найти вероятность P(|X - E(X)|<= sigma).

2. Дана функция распределения вероятностей случайной величины Х: F(X)=(3x^2+x)/52, 0<=x<=4

Найти плотность распределения вероятностей, математическое ожидание, дисперсию и
среднеквадратическое отклонение случайной величины Х. Найти вероятность попадания Х в интервал (2; 5).

3. На плоскости начерчены две окружности, радиусы которых 10 и 50 соответственно. Меньшая окружность содержится внутри большего круга. В большой круг наудачу бросаются 7 точек. Пусть случайная величина X — число точек, попавших в малый круг. Вычислите математическое ожидание E(X) и дисперсию D(X).

4. Случайные величины X1, . . . , X16 независимы и распределены по закону Пуассона с одинаковым математическим ожиданием, равным 8. Найдите математическое ожидание E(X1+ . . . +X16^2).

5. Случайная величина X имеет нормальное распределение с параметрами E(X) = 40 и дисперсией D(X)=sigma^2. Найдите вероятность попадания X в интервал (40-2sigma; 40).

Autmn term \ Осенний семестр

Statistics\ Статистика

1. In order to estimate the average intelligence quotient (IQ) of the students of a large university, the mean IQ of a number of randomly chosen students is calculated.
Using the central limit theorem, under the assumption that the distribution of the mean IQ is close to a normal one, estimate the minimal number of chosen students needed to obtain a result differing from the real average IQ by not more than 1 with a probability not less than 0.95, if it is known from previous studies that the standard deviation of IQ doesn't exceed 10. Estimate of the minimal number of chosen students.

2. For a sample of size 180 drawn from a normal population the unbiased point estimate of the population variance 0.5768 is found. Find a confidence interval (CI) for the population variance at confidence level 0.97.

(1−𝛾)/2 (or α/2-quantile); (1+𝛾)/2 (or (1–α/2)-quantile);
Left boundary of CI for the variance; Right boundary of CI for the variance.

3. A data set is given: {1.02, -0.7, -1.61, -1.59, 1.24, 0.91, 0.69, 0.58, 1.18, 0.13, -0.46, -0.21, 0.2, 2.05, -2.2, 1.39, -0.88, 0.9, 0.18, 1.13, -2.59, -1.18, 0.3, -0.81, -1.89, 0.42, -2, 0.87, -0.41, -0.8, 0.62, -1.34, -0.52, -0.21, -1.88, -0.66, -1.51, 1.46, -0.73, 1.33, 0.06, 0.01, 2.24, -0.77, 0.26, -0.04, -1.33, 1.88, 0.38, 0.33, 1.37, 0, -1.78, 0.3, 1.31, -0.64, 0.14, 0.53, -1.5, -1.02, -0.71, 0.75, 1.99, -0.94, -1.34, -0.21, 1.37, 0.9, -2.84, -0.08, 1.79, -0.35, -0.63, -2.09, -2.79, -0.11, -0.14, -1.52, -1.77, 0.58, -0.9, -1.68, 0.22, 0.46, 0.5, 1.19, -0.78, 0.02, -1.11, -0.58, 0.18, 0.11, 0.18, -0.9, 0, 0.11, 0.49, 0.01, -0.31, 1.89, -0.35, -0.21, 1.6, 0.59, -2.56, -0.42, -0.48, -0.46, 0.85, 0.39, -0.25, 0.86, 1, 0.87, 0.2, 1.69, -0.38, 1.62, 1.67, 0.34, -1.41, -0.07, 0.4, -1.04, -0.08, 0.56, 1.69, -1.44, -0.88, -1.35, -1.3, 0.54, -1.19, -1.51, 1.24, -1.14, 0.81, -0.15, -0.83, 1.07, 0.17, 0.63, 0.07, -1.45, -1.79, 0.32, -2.85, -0.75, 0.63, -0.22, 0.73, -0.57, -0.2, -1.3, -1.48, 0.82, -0.21, -0.74, 0.07, 1.39, -0.35, -0.67, 1.48, 2.15, -0.02}. Under the assumption that this data set is a sample drawn from a normal population compute a confidence interval (CI) for the population mean at confidence level 0.97.

4. A data set is given: {-1.33, -0.02, 1.45, 2.38, -0.55, 1.09, 0.21, -0.29, 0.61, !, 7.1234$, 0.83, 0.22, -0.32, -1.08, 1.09, NA, 0.69, 1.23, 0.49, 0.45, 0.33, 1.38, NA, 0.51, -0.15, 0.18, 0.2, -2.88, -1.42, 1.43, 1.99, -0.72, -2.49, -0.18, 1.68, 0.37, 0.24, 1.77, (, 2.01, 0.28, !0/0001, -0.92, -2.79, 1.64, 0.43, 0.79, -0.6, -0.07, 0.87, -0.44, -1.69, -1.93, -0.7, ), 1.4, 0.62, 2.15, 0.32, NA, -0.45, 0.08, -1.33, -0.9, -2.25, 0.07, -0.76, -0.68, -0.38, 0.78, 2.62, !., -0.75, 0.71, 1.*!@#$, -0.03, NA, -1.1, -0.72, 1.06, 0.05, 0.5, 0.9, -0.57, 17.*, -1.12, 1.37, -1.77, 0, -0.27, -0.74, 1.43, 1^.11243, 0.71, 0, 1.83, 0.96, 0.02, 3874.%!, 0.03, -2.2, 0.66, 0.32, -0.57, 0.58, $, 0.32, -1.57, 1.06, 1.32, -2.26, 0.72, 0.6, 0.02, 0.43, 0.05, 0.09, 0.81, -0.39, 0.24, -0.95, 0.83, -0.24, !.454345, -0.66, -0.31, 0.52, (3%:)?!:, 1.84, 1.55, -0.08, -0.19, 2.06, 3.08, 0.52, 0.52, -0.07, -0.28, -1.04, 0.32, -0.42, -0.21, 1.18, 1.67, -0.05, 0.15, -0.2, -0.95, -0.51, -0.35, 0.34, 0.39, 1.52, 0.13, -0.41, 0.44, ^, 0.24, -0.59, -0.07, -0.26, 0.57, 0.58, 0.75, 0.62, 0.12, 0.55, -0.14, -1.51, 0.87, 15&, -0.08, 0.65, -0.67}.

1. Using Excel and/or R(RStudio), clear this data set out of non-numeric values and omissions and compute the following statistical characteristics of the sample, cleared out of non-numeric values and omissions:
1.1.   Size of the sample, cleared out of non-numeric values and omissions
1.2.   Interquartile range
1.3.   Sample standard deviation
1.4.   Lower boundary of normal range
1.5.   Total number of outlying cases

2. In addition clear the sample out of outlying cases and compute the following statistical characteristics of the sample, cleared out of non-numeric value, omissions and outlying cases:
2.1   Size of the sample, cleared out of non-numeric values, omissions and outlying cases
2.2.   0.84-quantile
2.3.   Kurtosis (unbiased estimate)

5. For a sample of size = 42 sample moments are calculated:
expected value = 0.82, variance = 3.32
Using the method of moments find estimates of the parameters of a normal distribution. For one-parameter distributions use the first raw moment. For two-parameter distributions use additionally the second central moment.

1. Используя Excel и/или R(RStudio), проведите очистку набора от нечисловых значений и пропусков и вычислите статистические характеристики очищенной выборки: количество нечисловых значений и пропусков в исходной выборке, медиана, стандартное отклонение (несмещенное), нижняя граница нормы, верхняя граница нормы.

2. Дополнительно очистите выборку от выбросов и вычислите следующие статистические характеристики выборки, очищенной от нечисловых значений, пропусков и выбросов: объем выборки, очищенной от нечисловых значений, пропусков и выбросов, квантиль уровня 0.62, эксцесс (несмещенная оценка).

3. Методом моментов найдите оценки параметров равномерного распределения. Для однопараметрических распределений используйте первый начальный момент ν1. Для двухпараметрических распределений дополнительно используйте второй центральный момент μ2.

4. По извлеченной из нормальной генеральной совокупности случайной выборке объёма 152 найдена точечная несмещенная оценка дисперсии 1.3889. Постройте доверительный интервал (ДИ) для дисперсии на уровне доверия 0.96. Введите: квантиль уровня (1−𝛾)/2, квантиль уровня (1+𝛾)/2, левая граница ДИ для дисперсии, правая граница ДИ для дисперсии

5. Для оценки средней суммы чека в заведении общепита вычисляют среднее арифметическое некоторого количества случайно выбранных чеков. Используя центральную предельную теорему, в предположении, что распределение средней суммы чека не очень отличается от нормального, оцените минимальное количество выбранных чеков, необходимое для того, чтобы с вероятностью не менее 0.9 полученное значение отличалось от истинной средней суммы чека не более чем на 50 руб., если из предыдущих наблюдений известно, что среднеквадратичное отклонение суммы чека не превышает 200 руб.

Midterm \ Контрольная работа № 1

1. The time T (in hours) of failure-free operation of the company's server follows an exponential distribution with the parameter λ = 0.06. Find the probability P that the server will work without failures for no more than 27 hours. In the answer field, enter the value of the resulting probability P

2. Sports goods are supplied to the store by three supplier companies. The first supplier provides 42% of the total product, the second – 42% and the third – 16%. It is known that the percentage of defective goods obtained from the first company is equal to 1.5%, the second – 0.8% and from the third – 0.5%. What is the probability P that the randomly taken good turned out to be defective? Specify the most likely supplier (1,2 or 3) who provided this product.

3. The probability mass function for some discrete random variable X is given as follows:

X: -4, -2, 0, 2, 3. P: 0.05, 0.1, 0.05, 0.15, 0.65. Find for this random variable X its expected value E(X) and probability P(|X−E(X)|>σ(X)), whereas σ(X) stand for standard deviation.

4. A twelve-digit number is selected at random. Find the probability P that the number is read equally both from left to right and from right to left (such as the number 5328235).

5. In an insurance agency, a total revenue X (in million rubles) from sales of annual insurance contracts has a normal distribution with parameters m=97.8 and σ2=8.5, and total insurance payments for the year are described by the value Y, where Y is a random variable following the normal distribution with parameters m=31 and σ2=4.1 . The coefficient of correlation between revenue and insurance payments is ρ(X,Y)=0.37. Find the expected value and variance of the insurance agency's annual profit.

6. The probability density function f(x) of a continuous random variable X is given by the formula
f(x)={Cx6,x∈[−4;5],0,x∉[−4;5]. Find the constant C and expected value E(X).

Final Test \ Зачет

Probability Theory\ Теория вероятности

Test lasts 60 minutes. To enter Windows, please use login: studmoodle and password: moodle38. To enter campus.fa.ru use your regular login and password. Open RStudio and Excel before you start the test. Check if the Analysis ToolPak is at the Data panel. Name your file as “GROUP_LastName_FirstName"

Test lasts 60 minutes. To enter Windows use login: studmoodle and password: moodle38. To enter campus.fa.ru use your regular login and password. Open RStudio and Excel before you start the test. Name your file as “GROUP_LastName_FirstName"

Test 1 covers the following topics: Combinatorics, Classical and Geometrical Probability, Axioms of Probability, the Total Probability Formula, Bayes' Theorem, and General Types of Discrete Distributions.

1. Student knows 58 from 70 exam questions. What is a probability P of correct answer to 2 from 3 exam questions randomly chosen by an examiner.

2. 2 points are thrown randomly into a circle with a radius 130 Find the probability that a distance from a the centre of the circle to the nearest point will be not less than 78.

3. Probabilities of the following events are given: P(A)=0.77P, P(B)=0.86, P(C)=0.78. Find the minimum possible probability of the event A∩B∩C.

4. An owner of sound recordings shop has decided to classify his potential buyers in marketing research according to their ages, choosing categories "senior high school student", "colleges student" and "middle-aged and aged peoples". It was found that these three categories are presented in proportions 40%, 40% and 20%. It was also found, that purchases were made by 10% of senior high school students, by 35% of college students and by 15% of middle-aged and adged buyers. What is the probability P1 of an event: randomly chosen shop visitor will not make a purchase? If the randomly chosen shop visitor did not make a purchase, what is the probability P2 that she/he was senior high school student?

5. The probability to produce defective item is 0,03. Find the probability P that there are more than 384 items without defectives among 400 produced.

1. Discrete random variable X follows a distribution given as follows X: -4,-3,-1, 2, 3. P: 0.1, 0.15, 0.1, 0.15, 0.5. Find expected value E(X), standard deviation σ(X) and probability P(|X−E(X)|≤σ(X)).

2. A cumulative distribution function F(x) for a continuos random variable X takes a form F(x)={0,if x<4, (x^2−4x)/117, if 4≤x≤13, and 1, if x>13}. Find a probabilty P(6<X<10), percentile q=0.09 for a level 0.09 and standard deviation σ(X) for this random variable.

3. Random variables X1,…,X9 are independent and follow an exponential distribution with the same expected value equal 10. Find the following numerical characteristics: E(X1+⋯+X9−17),σ(X1+⋯+X9−17) and E[(X1+⋯+X9−17)^2].

4. A random variable Z is a linear combination of independent random variables X and Y: Z=0.81X+0.19Y, where a random variable X follows a normal distribution with expected value E(X)=1.9 and standard deviation σ=1.3, and a random variable Y follows a uniform distribution over the range [−26;22]. Find E(Z), Var(Z) and P(X≤6,Y>−10).

5. A random variable X follows a normal distribution with parameters m=8 and σ2=4.3, and a random variable Y follows a Poisson distribution with expected value E(Y)=3.6. A coefficient of correlation between X and Y is ρ(X,Y)=0.6. Find covariance Cov(X,Y), expected value E(3XY+11) and variance Var(4X−3Y+102).

Spring term \ Весенний семестр

End Semester Test 2 \ Контрольная работа № 2

1. Test the null hypothesis H0: Var(X)=Var(Y)H0:Var(X)=Var(Y) at significance level α=0.06α=0.06 against the alternative H1: Var(X)≠Var(Y)H1: Var(X)≠Var(Y), computing the P-value.

X={-0.37; 0.93; 0.29; 2.02; -1.29; -0.9; -0.06; 0.57; 0.84; 3.01; 1.1; -1; -1.26; -0.17; -2.41; 1.25; -0.49; 0.16; -1.33; 0.24; -0.31; 0.27; 0.9; 0.48; 1.09; 0.33; 0.53; 1.08; 0.66; 0.7; -0.09; -0.12; 0.47; 1.14; 0.47; 0.01; -0.63; -0.45; 1.35; -0.04; -1.08; 3.55; -2.52; -1.53; 0.53; 0.79; 1.84; 0.56; 1.35; -1.14; 0.11; 0.04; 1.11; 1.26; -0.19; 1.62; 0.23; 0.48; -1.95; -0.82; 0.78; 0.25; -2.19; 1.17; 0.87; 0.96; 0.16; -1.21; 2; 1.32; -2.06; 1.2; -0.55; 3.07; 1.39; 0.94; 0.22; 1.22; -0.11; 2.59; 1.1; 0.65; -2.32; 1.7; 1.04; 1.6; 1.38; -0.63; -0.98; -1.07; -0.87; 1.92; 0.43; -0.65; 0.68; -0.17; 3.57; 0.63; 0.56; 0.01; -1.06; 0.52; -0.42; -0.06; 0.87; 0.46; 1.33; -1.84; -0.57; 0.36; -0.13; 0.38; -1.3; -0.46; 0.7; 1.48; 0.91; 0.95; 0.33; 1.79; 1.62; -0.21; -0.48; 0.05; 0.46; 1.22; 0.23; 1.1; 1.25; -0.52; 0.65; 0.83; -0.47; 0.47; -0.85; -0.65; 1.14; -1.24; 2.32; 1.05; 3.02; 1.17; 1.61; 0.41; -1.62; 2.21; -0.6; 1.52; 0.83; 1.39; -0.38; -0.61; 1.15; 0.12; -0.16; -0.68; 0.48; 0.15; -0.01; 0.67; 0.35; 0.39; 1.02; 3.21; -0.34; -1.07; 0.37; -0.35; -1.55; -0.39; -0.24}

Y={0.61; -0.94; 1.7; 1.03; -0.99; 3.13; 2.15; 1.26; 0.44; 1.63; -1.93; -0.41; 2.27; 0.9; 1.88; 0.94; 1.2; -0.47; 0.53; 1.04; -1.09; 1.61; -0.01; 1.17; 0.36; -2.28; 0.02; -0.99; 1.99; 2.01; 0.01; -1.22; -0.1; 2.21; -0.31; -1.69; -0.75; 0.47; 1.35; -1.81; -0.17; 0.3; 1.49; 0.3; -0.21; 1.38; 2.76; 0.82; 1.63; 0.69; 1.3; 1.36; -1.93; 1.08; -0.97; 1.15; 0.43; 0.09; -3.13; 1.46; 0.88; -2.49; 1.69; -1.56; 0.95; 0.74; 1.93; 0.95; 0.63; -0.26; -0.57; 0.67; 1.46; 1.88; 3.16; -0.31; 2.06; -0.46; 1.41; 3.05; 1.01; 1.88; 1.31; -1.46; 2.17; -1.54; -1.08; 1.21; -0.45; 4.03; 1.06; 1.76; 0.2; 1.13; 0.42; 0.14; 0.68; -0.9; 0.27; 0.81; 0.77; -1.2; 2.1; 0.87; 0.88; 2.84; 1.17; 2.25; 1.5; 0.19; 1.1; 0.69; -1; 1.38; 1.44; -1.47; 1.12; -1.46; 0.87; -0.15; 2.7; 0.6; 1.11; 0.32; -0.09; 1.44; -0.79; 0.42; -0.48; 2.06; 0.1; 0.16; 1.79; 0.88; -1.8; 2.75; 0.4; 0.39; -0.03; 2.39; 1.45; 2.08; 2.36; 1.26; 0.24; 0.33; -0.13; 0.66; -0.05; 0.62; -1.07; -0.9; 1.63; -0.03; 3.82; -0.11; 1.48; -0.58; -1.08; -1.88; 0.56; 2.3; 1.75; 0.82; -0.26; -0.62; 0.89}

2. Using Excel or R(RStudio), clear the sample out of omissions, denoted as "NA". After that test the hypothesis about discrete uniform distribution of respondents' answers at significance level 0.09, applying goodness-of-fit (Pearson's chi-squared) test.
{Uneducated; Uneducated; High_School; NA; Uneducated; Doctorate; Unknown; Unknown; Unknown; Uneducated; Unknown; High_School; High_School; Graduate; Post-Graduate; NA; Post-Graduate; Uneducated; College; College; Uneducated; Uneducated; NA; Uneducated; Doctorate; Post-Graduate; Doctorate; Post-Graduate; Unknown; Unknown; High_School; Uneducated; High_School; Post-Graduate; Doctorate; High_School; Uneducated; Graduate; College; NA; Uneducated; Uneducated; Doctorate; Post-Graduate; NA; Post-Graduate; Unknown; College; Post-Graduate; Uneducated; Graduate; Doctorate; Uneducated; Post-Graduate; Uneducated; Unknown; NA; Uneducated; Doctorate; Doctorate; Post-Graduate; Doctorate; Uneducated; Post-Graduate; College; High_School; Doctorate; NA; High_School; Post-Graduate; Graduate; Doctorate; High_School; Graduate; Graduate; College; Unknown; High_School; NA; Unknown; Uneducated; Doctorate; College; College; Uneducated; Graduate; Doctorate; College; NA; Doctorate; Post-Graduate; NA; College; Unknown; Uneducated; College; Graduate; Graduate; Unknown; Uneducated; Unknown; Doctorate; Doctorate; Uneducated; Unknown; NA; NA; NA; Post-Graduate; Doctorate; NA; Unknown; NA; Uneducated; High_School; NA; High_School; Graduate; College; Doctorate; High_School; High_School; Post-Graduate; Uneducated; Post-Graduate; NA; Doctorate; Graduate; NA; NA; Uneducated; Doctorate}.

3. A two-dimensional sample (X,Y) is given. Delete all rows, in which at least one value is missing (omissions are denoted by "NA"). Test the hypothesis about insignificance of correlation coefficient ρ
(i.e. H0: ρ=0 against the alternative H1:ρ≠0)

To complete the task please check additional presentation at the page Statistics.

{(218.72,181.7728); (188.2,177.1048); (162.9,193.3612); (200.99,193.3356); (201.26,180.1644); (185.12,197.9724); (182.44,182.0812); (181.26,197.6728); (154.38,194.3908); (175.15,174.9292); (154.91,NA); (166.35,194.9528); (140.65,179.0508); (164.39,181.732); (199.24,218.0448); (167.96,186.3968); (175.74,183.3668); (NA,NA); (159.11,176.1024); (193.9,180.4036); (157.16,165.486); (192.68,185.55); (198.1,211.8816); (170.6,173.452); (173.83,204.3188); (177.47,165.8872); (217.79,164.7428); (152.67,188.3936); (155.88,183.8756); (196.24,195.9456); (139.69,181.6236); (195.38,186.962); (176.38,190.8056); (174.55,165.5892); (196.66,NA); (217.74,160.4424); (185.63,193.1096); (183.33,185.1884); (177.97,178.0896); (NA,NA); (158.6,NA); (183.66,153.53); (130.17,160.4472); (171.77,181.8808); (NA,NA); (177.62,190.592); (179.19,182.2628); (182.85,180.1084); (172.39,172.0864); (165.36,183.8336); (168.42,189.4788); (175.91,193.7672); (157.73,175.5228); (176.02,NA); (153.56,147.718); (165.73,174.2492); (173.85,198.0092); (164.51,167.4356); (174.33,190.8164); (199.74,147.5576); (183.78,168.2688); (NA,NA); (172.21,169.4776); (157.5,165.2464); (NA,NA); (156.27,NA); (184.19,199.232); (160.01,201.6676); (190.97,187.8328); (177.55,157.3928); (188.32,178.66); (194.08,157.234); (170.33,187.9296); (174.69,179.9708); (212.38,181.6704); (197.33,170.0796); (166.19,200.184); (162.97,179.5484); (193.07,184.8728); (189.09,192.7608); (150.08,NA); (161.17,182.2656); (177.4,189.1024); (191.79,NA); (179.76,NA); (181.59,188.26); (205.14,180.99); (198.27,213.1464); (164.82,178.0404); (189.87,194.5988); (169.94,179.7932); (175.1,200.4092); (191.38,175.602); (160.65,168.2584); (162.02,204.4872); (173.01,159.5412); (182.28,183.4392); (180.74,NA); (154.47,177.0468); (NA,NA); (165.57,175.4232); (163.52,197.192); (186.44,180.3956); (164.11,174.3404); (152.15,NA); (179.57,174.0408); (188.72,185.5368); (172.29,NA); (211.68,223.318); (156.08,198.8692); (NA,NA); (176.49,201.6428); (171.92,NA); (233.06,179.654); (174.47,172.124); (179.67,200.9772); (156.13,191.0992); (226.14,198.8436); (201.65,154.454); (191.79,NA); (142.49,NA); (186.2,197.2492); (165.99,207.528); (181.67,182.866); (153.76,188.858); (157.73,168.0616); (179.95,162.3044); (175.21,194.576); (158.44,134.7592); (159.4,190.128); (162.51,171.8664); (192.52,193.21); (192.65,181.7204); (160.83,174.1976); (188.16,194.2964); (180.38,185.0812); (195.96,170.6416); (185.85,NA); (168.03,185.2248); (173.54,188.3428); (178.7,211.8016); (229.25,167.6192); (156.08,168.9508); (170.8,224.4544); (187.64,207.1808); (153.96,173.096); (157.35,186.5508); (180.57,224.224); (162.32,207.7772); (153.6,179.544); (191.78,177.2072); (173.67,176.7704); (176.56,173.0916); (NA,NA); (169.18,147.7164); (189.74,190.936); (192.84,NA); (NA,NA); (198.72,188.7012); (NA,NA); (187.41,185.8276); (188.25,178.8016); (132.99,192.6152); (214.12,198.756); (184.21,NA); (184.91,193.2728); (166.09,162.9436); (170.54,167.7156); (174,NA); (NA,NA); (162.56,176.9488); (183.14,144.7208); (170.19,187.7804); (202.5,196.198); (NA,NA); (171.35,NA); (150.73,183.114); (174.42,213.0876); (191.44,199.1956); (183.25,219.7556); (186.56,169.3744); (174.85,180.1676)}.

4. Two samples are drawn from two normal populations X and Y. Test the null hypothesis H0: E(X)=E(Y) (without the assumption of equality of variances) at significance level α=0.09 against the alternative H1: E(X)≠E(Y), computing the P-value.
X={-3; -0.3; -0.59; -1; -0.28; -2.02; 0.34; 0.52; -0.86; -1.01; -2.49; -2.2; -0.72; 2.37; -0.06; -2.54; -1.17; 1.47; 0.6; -0.38; -1.45; 1.9; -0.22; 3.77; -2.33; -0.77; 1.79; 0.85; -2.49; -1.23; -0.26; -1.71; -0.31; 1.45; 0.87; 0.97; -1.83; 1.12; -0.54; 0.86; 2.67; 0.32; 1.13; 0.88; 0.22; -1.81; -0.15; -0.45; -2.85; -0.77; -0.01; -0.95; 1.33; 1.33; -0.09; 1.38; 4.85; -0.5; -2.32; 1.33; -0.14; -0.99; -1.66; -1.09; 1.35; 1.84; 1.5; -0.01; 0.75; 2.21; 2.21; 0.41; -1.85; -1.93; -2.26; -0.82; -0.86; 1.99; 0.79; -2.25; 0.13; -0.34; -2.02; -3.36; 1.76; -1.46; -0.32; -1.08; -2.55; 0.16; 2.75; -0.45; 2.16; 1.82; 1.27; -0.33; -0.13; -4.14; 0.13; 0.75; -0.89; -1.93; -1.43; 0.72; 0.99; 1.54; -0.58; -1.13; 3.17; -1.6; 0.89; -2.74; 0.04; 0.51; -2.26; -1.96; -0.03; 0.31; -2.52; 0.24; -0.25; 1.67; 0.49; -1.46; 0.07; -3; 2.16; -0.15; -0.35; 0.86; -2.75; 0.69; 0.82; 0.5; -2.03; 0.01; -2.11; -0.65; 0.93; -0.4; -1.4; 1.95; -1.21; -1.13; 1.19; -0.57; 0.68; -1.98; -0.71; -0.62; -0.69; -0.23; 1.69; -1.19; 1.02; 0.45; -1.16; 1.78; 0.37; -1.16; -0.52; -0.56; -0.14; 0.29; -1.72; -1.29; 0.3; 1.57; -1.33; -1.77; -1.53; -1.76; -0.36}
Y={0.84; 0.72; 1.75; 1.55; -1; -0.06; 0.96; 0.49; 0.53; -0.79; 1.21; 2.55; 1.71; 2.37; -0.03; -1.28; -0.32; 0.06; -1.36; 1.68; 0.49; 1.51; -0.38; 2.38; 0.39; -0.3; -0.38; 0.53; 0.72; 0.23; -0.31; 1.35; -0.59; 0.28; -0.63; -1.12; -1; 0.26; 2.21; -0.79; 2.25; -0.27; 2.12; -1.67; -1.32; 0.81; 0.96; -1.21; 0.47; -0.26; 0.81; -0.92; 0.41; 0.1; -0.06; 0.59; 1.5; -0.04; 0.26; 3.31; 2.39; -2.1; 0.33; 0.17; 0.79; -2.36; 1.25; 1.77; -1.61; 0.79; 0.42; -1.46; 0.35; 0.43; 1.48; 1.75; 1.92; 1.86; 1.31; -0.01; 1.41; -0.81; 2.01; -0.43; 2.13; -1.46; -1.35; 0.18; 0.45; -0.42; 0.23; -1.73; -0.76; -0.95; 2.16; -1.03; 0.96; -0.19; -2.49; -0.9; -0.27; -0.7; -0.7; 2.11; 0.6; -2.35; 1.38; 0.04; -0.57; 0.72; 2.27; 1.27; 0.26; 0.58; 0.64; -0.58; 0.46; 1.47; 0.65; 0.68; 1.8; -0.25; -0.59; 1.83; 0.24; 1.51; -1.17; 0.52; -0.65; -1.58; 1.18; -1.33; 1.15; 0.58; 0.55; 2.75; 2.39; -0.98; -1.72; 0.84; 1.05; -1.21; 1.84; 0.81; 2.03; 0.38; -1; -0.64; 1.72; -0.62; 1.52; -0.88; 0.29; 2.38; 2; 1; 0.48; 0.28}

5. According to the results of a sociological study, respondents' answers to a survey questions are represented as a sample. Using Excel or R(RStudio), clear the sample out of omissions, denoted as "NA", and answer the following questions:
1. Enter the number of different variants of respondents' answers occurring in the cleared sample
2. Enter the number of respondents who gave answer "Post-Graduate"
3. Enter the proportion of respondents who gave answer "Post-Graduate"
4. Enter the left boundary of a 0.92-confidence interval for the population proportion of answers "Post-Graduate"
5. Enter the right boundary of a 0.92-confidence interval for the population proportion of answers "Post-Graduate".
{Unknown; Post-Graduate; High_School; High_School; Uneducated; High_School; Graduate; Graduate; Graduate; Post-Graduate; NA; Post-Graduate; Uneducated; Unknown; Post-Graduate; NA; NA; Post-Graduate; Uneducated; Unknown; NA; High_School; NA; Graduate; Uneducated; High_School; College; Graduate; High_School; NA; Unknown; High_School; NA; Uneducated; High_School; Post-Graduate; Unknown; Post-Graduate; Graduate; Post-Graduate; Graduate; Graduate; NA; Uneducated; Post-Graduate; Unknown; Post-Graduate; NA; NA; Post-Graduate; Graduate; Graduate; College; Post-Graduate; High_School; Post-Graduate; Uneducated; Uneducated; Uneducated; Unknown; Unknown; High_School; Unknown; NA; High_School; Uneducated; NA; Graduate; High_School; Uneducated; College; High_School; Graduate; Uneducated; Post-Graduate; Post-Graduate; High_School; Unknown; Post-Graduate; High_School; Post-Graduate; Uneducated; Graduate; High_School; Graduate; NA; High_School; Post-Graduate; Post-Graduate; NA; Unknown; College; NA; Unknown; Unknown; Post-Graduate; College; NA; Unknown; College; High_School; College; College; Uneducated; College; Post-Graduate; Graduate; Uneducated; Uneducated; High_School; Graduate; College; High_School; High_School; High_School; High_School}.

To enter Windows, please use login: studmoodle and password: moodle38. To enter campus.fa.ru use your regular login and password. Open RStudio and Excel before you start the test. Name your file as “GROUP_LastName_FirstName"

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To read the sequences, please try to use scan("clipboard", what = character()) or scan("clipboard", what = numeric()) or read from Excel file^ library(readxl) read_excel("path")