Statistics Questions

Consider the following distribution of 50 workers of a factory. Find the mean daily wages of the workers of the factory by using an appropriate method.

Question : Daily Wages in Rupees 0 – 120 120 – 140 14 – 160 160 – 180 180 – 200 Number of Workers 12 14 8 6 10 Solution : Let the assumed mean A = 150 and class size h = 50 So, \(u_i\) = \(x_i – A\over h\) = \(x_i – 150\over …

Consider the following distribution of 50 workers of a factory. Find the mean daily wages of the workers of the factory by using an appropriate method. Read More »

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Question : Number of plants 0 – 2 2 – 4 4 – 6 6 – 8 8 – 10 10 – 12 12 – 14 No. of houses 1 2 1 5 6 2 3 Which method do you use for finding the mean and why ? Solution : Number of Plants Number of …

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house. Read More »

In a class of 100 students. there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class is 72, then what is the average of the girls?

Solution : Since, total number of students = 100 and number of boys = 70 \(\therefore\) number of girls = (100 – 70) = 30 Now, the total marks of 100 students = 100*72 = 7200 And total marks of 70 boys = 70*75 = 5250 Total marks of 30 girls = 7250 – 5250 …

In a class of 100 students. there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class is 72, then what is the average of the girls? Read More »

The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X = 1) is

Solution : Given that, for binomial distribution mean, np = 4 and variance, npq = 2. \(\therefore\)  q = 1/2, but p + q = 1 \(\implies\) p = 1/2 and n \(\times\) \(1\over 2\) = 4 \(\implies\) n = 8 We know that,  P(X = r) = \(^nC_r p^r q^{n-r}\) \(\therefore\)  P(X = 1) …

The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X = 1) is Read More »

The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observation of the set is increased by 2, then the median of the new is

Solution : Median of new set remains the same as of the original set. Similar Questions The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X = 1) is Let \(x_1\), \(x_2\), ….. , \(x_n\), be n observations such that \(\sum{x_i}^2\) = 400 and \(\sum{x_i}\) …

The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observation of the set is increased by 2, then the median of the new is Read More »

The mean and the variance of a binomial distribution are 4 and 2, respectively. Then, the probability of 2 success is

Solution : Given that, mean = 4 \(\implies\) np = 4 And Variance = 2 \(\implies\) npq = 2 \(\implies\) 4q = 2 \(\implies\)  q = \(1\over 2\) \(\therefore\)   p = 1 – q = 1 – \(1\over 2\) = \(1\over 2\) Also, n = 8 Probability of 2 successes = P(X = 2) = …

The mean and the variance of a binomial distribution are 4 and 2, respectively. Then, the probability of 2 success is Read More »

In a series of 2n observations, half of them equals a and remaining half equal -a. If the standard deviation of the observation is 2, then |a| equal to

Solution : In the 2n observations, half of them equals a and remaining half equal -a. Then, the mean of total 2n observations is equal to 0. \(\therefore\)   SD = \(\sqrt{\sum(x – \bar{x})^2\over N}\) \(\implies\)  4 = \(\sum{x^2}\over 2n\) \(\implies\)  4 = \(2na^2\over 2n\) \(\implies\)  \(a^2\) = 4 \(\therefore\)   a = 2 Similar Questions The …

In a series of 2n observations, half of them equals a and remaining half equal -a. If the standard deviation of the observation is 2, then |a| equal to Read More »

A random variable X has poisson distribution with mean 2. Then, P(X > 1.5) equal to

Solution : Now, P(X > 1.5) = P(2) + P(3) + …… \(\infty\) = 1 – [P(0) + P(1)] = 1 – \((e^{-2} + {e^{-2}(2)\over 1})\) = 1 – \(3\over e^2\) Similar Questions The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X = 1) …

A random variable X has poisson distribution with mean 2. Then, P(X > 1.5) equal to Read More »

If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately

Solution : Given that, mean = 21 and median = 22 Using the relation, Mode  = 3 median – 2 mean \(\implies\) Mode = 3(22) – 2(21) = 66 – 42 = 24 Similar Questions The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X …

If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately Read More »

Let \(x_1\), \(x_2\), ….. , \(x_n\), be n observations such that \(\sum{x_i}^2\) = 400 and \(\sum{x_i}\) = 80. Then, a possible value of among the following is

Question : Let \(x_1\), \(x_2\), ….. , \(x_n\), be n observations such that \(\sum{x_i}^2\) = 400 and \(\sum{x_i}\) = 80. Then, a possible value of among the following is (a) 12 (b) 9 (c) 18 (d) 15 Solution : Given  \(\sum{x_i}^2\) = 400 and \(\sum{x_i}\) = 80 \(\because\) \(\sigma^2\) \(\ge\) 0 \(\therefore\)  \(\sum{x_i}^2\over n\) – …

Let \(x_1\), \(x_2\), ….. , \(x_n\), be n observations such that \(\sum{x_i}^2\) = 400 and \(\sum{x_i}\) = 80. Then, a possible value of among the following is Read More »