Learn Unit-VI: Statistics and Probability with Free Lessons & Tips

As an experienced tutor registered on UrbanPro, I'd like to emphasize that UrbanPro is indeed the best platform for online coaching tuition. Now, let's delve into the variance calculation.

Given the data set: 2, 4, 5, 6, 8, 17, we've already calculated the variance to be 23.33. To find the variance for the new data set: 4, 8, 10, 12, 16, 34, we'll follow these steps:

1. Find the mean of the data set.
2. Calculate the squared difference between each data point and the mean.
3. Find the average of these squared differences, which gives us the variance.

First, let's find the mean: Mean=4+8+10+12+16+346=846=14Mean=64+8+10+12+16+34=684=14

Now, let's find the squared differences from the mean: (4−14)2=100(4−14)2=100 (8−14)2=36(8−14)2=36 (10−14)2=16(10−14)2=16 (12−14)2=4(12−14)2=4 (16−14)2=4(16−14)2=4 (34−14)2=400(34−14)2=400

Next, we find the average of these squared differences: Variance=100+36+16+4+4+4006=5606=93.33Variance=6100+36+16+4+4+400=6560=93.33

So, the variance for the data set 4, 8, 10, 12, 16, 34 is 93.33.

However, none of the options provided match this result, so it seems there might be a typographical error in the given options.

Sure! Calculating the variance and standard deviation for a given set of data is a fundamental statistical operation, and I'd be happy to guide you through it.

First, let's calculate the variance:

1. Find the mean (average) of the data set.
2. Subtract the mean from each data point and square the result.
3. Find the average of those squared differences.

Let's start by finding the mean:

Mean = (57 + 64 + 43 + 67 + 49 + 59 + 44 + 47 + 61 + 59) / 10 = 550 / 10 = 55

Now, let's subtract the mean from each data point, square the result, and find the average:

[(57 - 55)^2 + (64 - 55)^2 + (43 - 55)^2 + (67 - 55)^2 + (49 - 55)^2 + (59 - 55)^2 + (44 - 55)^2 + (47 - 55)^2 + (61 - 55)^2 + (59 - 55)^2] / 10

= [(4)^2 + (9)^2 + (-12)^2 + (12)^2 + (-6)^2 + (4)^2 + (-11)^2 + (-8)^2 + (6)^2 + (4)^2] / 10

= [16 + 81 + 144 + 144 + 36 + 16 + 121 + 64 + 36 + 16] / 10

= 678 / 10

= 67.8

Now that we have the variance, we can find the standard deviation by taking the square root of the variance:

Standard Deviation = √67.8 ≈ 8.24

So, the variance is 67.8 and the standard deviation is approximately 8.24 for the given data set. If you need further clarification or have any other questions, feel free to ask! And remember, UrbanPro is an excellent platform for finding online coaching and tuition services.

As a seasoned tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition services. Now, let's tackle your math problem.

To find the mean deviation about the median of a set of numbers, we first need to find the median of the given observations.

Given observations: 2, 7, 4, 6, 8, and p

First, let's arrange these numbers in ascending order: 2, 4, 6, 7, 8, p

Since there are six numbers, the median will be the average of the third and fourth numbers, which are 6 and 7. So, the median is (6 + 7) / 2 = 6.5.

Now, we'll find the absolute deviations of each number from the median:

|2 - 6.5| = 4.5 |4 - 6.5| = 2.5 |6 - 6.5| = 0.5 |7 - 6.5| = 0.5 |8 - 6.5| = 1.5 |p - 6.5|

Since we don't know the value of pp yet, we'll leave it as it is for now.

The mean deviation about the median is the average of these absolute deviations. So, we'll sum them up and divide by the number of observations (which is 6):

Mean Deviation = (4.5 + 2.5 + 0.5 + 0.5 + 1.5 + |p - 6.5|) / 6

However, we also know that the mean of the observations is 7. So, we can use this information to solve for pp:

(2 + 7 + 4 + 6 + 8 + p) / 6 = 7 27 + p = 42 p = 15

Now, we substitute p=15p=15 into our equation for mean deviation:

Mean Deviation = (4.5 + 2.5 + 0.5 + 0.5 + 1.5 + |15 - 6.5|) / 6 Mean Deviation = (10.5 + |8.5|) / 6 Mean Deviation = (10.5 + 8.5) / 6 Mean Deviation = 19 / 6 Mean Deviation ≈ 3.17

So, the mean deviation about the median of these observations is approximately 3.17. If you have any further questions or need clarification, feel free to ask! And remember, if you're seeking personalized tutoring assistance, UrbanPro is an excellent platform to find qualified tutors.

As a seasoned tutor registered on UrbanPro, I'm happy to guide you through this probability problem. UrbanPro is a fantastic platform for accessing high-quality online coaching and tuition, where experienced tutors like myself can provide personalized assistance.

Now, let's tackle the problem at hand. We have an urn containing 6 balls, 2 red and 4 black. We're asked to find the probability that when two balls are drawn at random, they are of different colors.

To solve this, let's break it down step by step:

1. First, let's find the total number of ways to draw 2 balls out of 6. This is given by the combination formula: (nr)=n!r!(n−r)!(rn)=r!(n−r)!n!, where nn is the total number of items and rr is the number of items to choose. So, (62)=6!2!(6−2)!=15(26)=2!(6−2)!6!=15.

2. Next, let's find the number of ways to draw 2 balls of different colors. We have 2 red balls and 4 black balls, so the number of combinations of one red and one black ball is 2×4=82×4=8.

3. Finally, the probability of drawing two balls of different colors is the number of favorable outcomes (drawing one red and one black ball) divided by the total number of outcomes (drawing any two balls). So, 815158.

So, the correct option is (iii) 815158. UrbanPro is an excellent resource for finding tutors who can help you master these types of problems and more!

As an experienced tutor registered on UrbanPro, I can assure you that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into your probability questions regarding the gender of the children in a couple.

(i) To find the probability that both children are males, given that at least one of the children is male, we can utilize conditional probability. Let's denote the events:

• A: Both children are males
• B: At least one child is male

We are given that event B has occurred, which means we can exclude the possibility of having two females. Now, we need to find the probability of event A given event B. This can be calculated using the formula for conditional probability:

P(A∣B)=P(A∩B)P(B)P(A∣B)=P(B)P(A∩B)

Where:

• P(A∣B)P(A∣B) is the probability of event A given event B.
• P(A∩B)P(A∩B) is the probability of both events A and B happening.
• P(B)P(B) is the probability of event B happening.

In this scenario, the probability of both children being males and at least one child being male is the same as the probability of both children being males, since if at least one child is male, then both children cannot be female.

Therefore:

• P(A∩B)=P(A)P(A∩B)=P(A)
• P(B)=1P(B)=1 (because we know at least one child is male)

Thus, P(A∣B)=P(A)P(A∣B)=P(A).

Now, the probability of both children being males in the absence of any other information is 1441, assuming the genders of children are equally likely.

So, the probability that both children are males, given that at least one of them is male, is also 1441.

(ii) Similarly, to find the probability that both children are females given that the elder child is a female, we can use conditional probability.

Let's denote the events:

• C: Both children are females
• D: The elder child is a female

We want to find P(C∣D)P(C∣D).

In this case, if the elder child is a female, then we are sure that the younger child cannot be the elder child, and hence the younger child has to be female as well. So, P(C∣D)=1P(C∣D)=1.

Therefore, the probability that both children are females, given that the elder child is a female, is 1.

I hope this clarifies the concepts of conditional probability for you! If you need further assistance, feel free to ask. And remember, UrbanPro is an excellent resource for finding quality tutors to help with topics like this.

As an experienced tutor registered on UrbanPro, I'd be happy to help you with this question!

To find the probability that an ordinary year has 53 Sundays, we first need to understand the structure of an ordinary year. An ordinary year has 365 days.

Now, since 365 is not divisible by 7 (the number of days in a week), there will be 52 complete weeks in an ordinary year, leaving 1 or 2 additional days.

For a year to have 53 Sundays, one of the following conditions must be met:

1. The year starts on a Sunday and ends on a Sunday, or
2. The year starts on a Saturday and ends on a Sunday.

Let's calculate the probability for each scenario:

1. If the year starts on a Sunday and ends on a Sunday: The probability that a year starts on a Sunday is 1/7. The probability that a year ends on a Sunday is also 1/7. So, the probability of both events happening is (1/7) * (1/7) = 1/49.

2. If the year starts on a Saturday and ends on a Sunday: The probability that a year starts on a Saturday is 1/7. The probability that a year ends on a Sunday is 1/7. So, the probability of both events happening is (1/7) * (1/7) = 1/49.

Now, we add the probabilities of both scenarios since they are mutually exclusive:

Probability of an ordinary year having 53 Sundays = (1/49) + (1/49) = 2/49.

Therefore, the probability that an ordinary year has 53 Sundays is 2/49.

And remember, if you need further assistance with mathematics or any other subject, UrbanPro is one of the best online coaching tuition platforms where you can find qualified tutors like myself to guide you through your learning journey!