# Free AIOU Solved Assignment Code 804 Spring 2023

## Free AIOU Solved Assignment Code 804 Spring 2023

Download *Aiou solved assignment* code 804 free autumn/spring 2023, aiou updates solved assignments. Get free AIOU All Level Assignment from aiousolvedassignment.

**Course: Statistics for Economist (804)**

**Semester: Spring, 2023**

**ASSIGNMENT No. 1**

**Q.1 **

**a) A student is given 4 questions, each with a choice of 3 answer. Let ‘X’ be number of correct answer when the students has to guess each answer. Compute the probability function and the mean and variance of ‘X’.**

P(x=3)

=5C_{3}

= (0.25)^{3}(0.75)^{2}

= 0.00878625

**b) Three coins are independently flipped; let X=number of heads. Make a table of other probability function, and find the probability of X, assuming that the coins are fair.**

Probability of getting head (success)=p=1/2

Probability of getting tail (failure) =1-p=1-(1/2) =1/2

Number of success=X

Since there are two mutually exclusive outcomes of independent trials (success or failure)

So X~Binomial Distribution with parameters n and p

pmf of Binomial =nCx * p^x *(1-p)^(n-x) {in this case n=3}

Probability Distribution of X is

X=0, P(X=0) =1/8

X=1, P(X=1) =3/8

X=2, P(X=2) =3/8

X=3, P(X=3) =1/8

## AIOU Solved Assignment Code 804 Spring 2023

**Q.2 **

**i) Two cards are drawn from an ordinary deck. What is the probability that:****They are black**

In a pack of 52 cards; there are total 26 black cards (cards belonging to club and spade suites)

Both are black cards = (26C2) = 325

**Red and black aces**

There are 52 cards in a deck, and there are 4 aces in a deck, so the probability of drawing an ace is 4 / 52.

If you do not replace the card, there are only 51 cards left and only 3 aces left, so the probability of now drawing an ace is 3 / 52.

Multiply these probabilities:

4 / 52 x 3 / 51=12 / 2652→0.00452

Therefore, the probability that both cards are aces is 1 / 13 ⋅ 1 / 13=1 / 169.

**They both queen**

4 queens (one each in club, heart, spade and diamond suites)

n (queen) = 4

Both are queens = (4C2) = 6

**King and Jack**

There are 4 kings in a deck of 52 cards (King of Hearts, King of Diamonds, King of Clubs & King of Spades)

Probability of getting a card that not a king

= 52 – 4

= 48

So there are 48 cards that are not King cards.

P(not a king)= 48/52 Or 12/13

A standard deck has 13 ordinal cards (Ace, 2-10, Jack, Queen, King) with one of each in each of four suits (Hearts, Diamonds, Spades, Clubs), for a total of 13×4=52 cards.

P(randomly draw a Jack)=16 / 52=4 / 13

**ii) In a family of 5 children, what is the chance?****At least one girl?**

# Probability of one girl = 1/4

**At least 3 boys, given at least one girl?**

Probability of at least one girl = 1/4 + 1/8 + 1/16 = 7 / 16

**At least 1 boy?**

# Probability of one Boy = 1 / 4

**At least 3 boys, given eldest is a boy?**

Probability of all three Boys = (1/4)*(1/4)*(1/4) = 1 / 64

## AIOU Solved Assignment 1 Code 804 Spring 2023

**Q.3 **

**a) The probability of snow tomorrow 0.7 and the probability that it will be colder is 0.6. The probability that it will not snow and not be colder is 0.1. What in the probability that I will snow if it colder tomorrow?**

Events:

C=colder; ~C=not colder

S=snow; ~S=not snow

P(S)=0.7

P(C)=0.6

P(~S∩~C)=0.1

by De Morgan’s law

P(S∪C)=1-P(~S∩~C)=1-0.1=0.9

=>

P(S∩C)=P(S)+P(C)-P(S∪C)=0.6+0.7-0.9=0.4

P(S|C)=P(S∩C)/P(C)=0.4/0.7=4/7=0.571 (approx.)

**b) Briefly explain the following:**

**(i) Sample space **

A sample space is a collection or a set of possible outcomes of a random experiment. The sample space is represented using the symbol, “S”. The subset of possible outcomes of an experiment is called events. A sample space may contain a number of outcomes which depends on the experiment. If it contains a finite number of outcomes, then it is known as discrete or finite sample spaces.

The samples spaces for a random experiment is written within curly braces “ { } “. There is a difference between the sample space and the events. For rolling a die, we will get the sample space, S as {1, 2, 3, 4, 5, 6 } whereas the event can be written as {1, 3, 5 } which represents the set of odd numbers and { 2, 4, 6 } which represents the set of even numbers. The outcomes of an experiment are random and the sample space becomes the universal set for some particular experiments. Some of the examples are as follows:

## Tossing a Coin

When flipping a coin, two outcomes are possible, such as head and tail. Therefore the sample space for this experiment is given as

Sample Space,S = { H, T } = { Head, Tail }

## Tossing Two Coins

When flipping two coins, the number of possible outcomes are four. Let, H_{1} and T_{1 }be the head and tail of the first coin and H_{2} and T_{2} be the head and tail of the second coin respectively and the sample space can be written as

Sample Space, S = { (H_{1}, H_{2}), (H_{1}, T_{2}), (T_{1}, H_{2}), (T_{1}, T_{2}) }

In general, if you have “n” coins, then the possible number of outcomes will be 2^{n}.

Example: If you toss 3 coins, “n” is taken as 3.

Therefore, the possible number of outcomes will be 2^{3} = 8 outcomes

Sample space for tossing three coins is written as

Sample space S = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

**(ii) Event **

In probability theory, an event is an outcome or defined collection of outcomes of a random experiment. Since the collection of all possible outcomes to a random experiment is called the sample space, another definiton of event is any subset of a sample space. For example, on the roll of a die, getting an even number is an event. This event is a subset containing sample points{2, 4, 6}. The sample space is{1, 2, 3, 4, 5, 6}

**(iii) Dependent and independent events **

For events to be considered dependent, one must have an influence over how probable another is. In other words, a dependent event can only occur if another event occurs first.

While this is a mathematic/statistical term, speaking specifically to the subject of probabilities, the same is true of dependent events as they occur in the real world.

For example, say you’d like to go on vacation at the end of next month, but that depends on having enough money to cover the trip. You may be counting on a bonus, a commission, or an advance on your paycheck. It also most likely depends on you being given the last week of the month off to make the trip.

The primary focus when analyzing dependent events is probability. The occurrence of one event exerts an effect on the probability of another event. Consider the following examples:

- Getting into a traffic accident is dependent upon driving or riding in a vehicle.
- If you park your vehicle illegally, you’re more likely to get a parking ticket.
- You must buy a lottery ticket to have a chance at winning; your odds of winning are increased if you buy more than one ticket.
- Committing a serious crime – such as breaking into someone’s home – increases your odds of getting caught and going to jail.

An event is deemed independent when it isn’t connected to another event, or its probability of happening, or conversely, of not happening. This is true of events in terms of probability, as well as in real life, which, as mentioned above, is true of dependent events as well.

For example, the color of your hair has absolutely no effect on where you work. The two events of “having black hair” and “working in Allentown” are completely independent of one another.

Independent events don’t influence one another or have any effect on how probable another event is.

Other examples of pairs of independent events include:

- Taking an Uber ride and getting a free meal at your favorite restaurant
- Winning a card game and running out of bread
- Finding a dollar on the street and buying a lottery ticket; finding a dollar isn’t dictated by buying a lottery ticket, nor does buying the ticket increase your chances of finding a dollar
- Growing the perfect tomato and owning a cat

**(iv) Mutually exclusive events **

**Mutually exclusive events**** are** things that can’t happen at the same time. For example, you can’t run backwards and forwards at the same time. The events “running forward” and “running backwards” are mutually exclusive. Tossing a coin can also give you this type of event. You can’t toss a coin and get both a heads *and* tails. So “tossing a heads” and “tossing a tails” are mutually exclusive. Some more examples are: your ability to pay your rent if you don’t get paid, or watching TV if you don’t have a TV.

## AIOU Solved Assignment 2 Code 804 Spring 2023

**Q.4 **

**a) Describe the central limit theorem in detail.**

The central limit theorem states that if you have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population **with replacement**, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large (usually n > 30). If the population is normal, then the theorem holds true even for samples smaller than 30. In fact, this also holds true even if the population is binomial, provided that min(np, n(1-p))> 5, where n is the sample size and p is the probability of success in the population. This means that we can use the normal probability model to quantify uncertainty when making inferences about a population mean based on the sample mean.

For the random samples we take from the population, we can compute the mean of the sample means:

and the standard deviation of the sample means:

Before illustrating the use of the Central Limit Theorem (CLT) we will first illustrate the result. In order for the result of the CLT to hold, the sample must be sufficiently large (n > 30). Again, there are two exceptions to this. If the population is normal, then the result holds for samples of any size (i..e, the sampling distribution of the sample means will be approximately normal even for samples of size less than 30).

If you’ve taken some calculus, you can define the CLT more precisely using the definition of a limit. The CDF of the standardized sample mean (X̄ – μ)/σ converges pointwise to the CDF (Φ) of the standard normal distribution. This is shown with the integral:

Where:

An assumption is that the expected value of X and X^{2} < infinity.** **

**b) The weights of packages filled by machine are normally distributed about a mean of 25 ounces, with a standard deviation of one ounce. What is the probability that n packages from the machine will have an average weight of less than 24 ounces if n = 1, 4, 16, 64?**

## AIOU Solved Assignment Code 804 Autumn 2023** **

**Q.5 On a certain Wednesday evening, a check was made of five different computer rooms in campus residence units. The number of students using computer in the five units was 23, 16, 34, 27 and 23 respectively.**

**a) Find the average number of users per room.**

Total users = 23+16+34+27+23 / 5

Total users = 24** **

**b) Find the variance of this sample distribution.**

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