Exercises for Calculating Normal Distribution and Probability

In this case, we are working with a standard normal distribution. To solve it we will use the following formula:

Now, we need to locate in our normal distribution table the value when , but we need the value for when , so we use , then we obtain . Additionally, since the normal distribution is symmetric, we have .

That is, approximately of the values of are within three standard deviations of the mean.

Using the formula , we will substitute the value of the mean () and the standard deviation ().

Simplifying, we obtain:

From which it follows that:

Now we locate in the normal distribution table the value and observe that it corresponds to , then:

Using the formula , we will substitute the value of the mean () and the standard deviation ().

We look up the corresponding values in the normal distribution table:

Therefore:

This means that in the entire month, only days will reach temperatures between and degrees Fahrenheit.

1. Between and .

Substituting:

Locating the values in the normal distribution table and operating:

Therefore, if we multiply the probability by the students:

Of the students, weigh between and pounds.

2. More than .

Substituting and simplifying:

Multiplying the probability by :

It is impossible to find a single student weighing more than pounds.

3. Less than .

Substituting and simplifying:

Multiplying the probability by :

There are students who weigh less than pounds.

4. .

When the distribution is continuous, the probability that the variable has an exact value is always zero (). Therefore:

5. or less.

Given the previous results:

There are zero students who weigh exactly pounds and there are students who weigh less than pounds, so there are students who weigh pounds or less.

1. What is the probability that a person taking the exam obtains a grade higher than ?

We substitute the values in the formula:

The probability that a person obtains a score greater than on the exam is .

2. Calculate the proportion of students who have scores that exceed by at least five points the score that marks the boundary between Pass and Fail (the of students who obtained the lowest scores are declared Fail).

Substituting values in the formula:

We locate the probability in the normal distribution table, which is , this means:

We solve for :

We calculate for :

The percentage of students who Pass and whose score is units above the Fail boundary is .

3. If it is known that a student's grade is greater than , what is the probability that their grade is, in fact, greater than ?

We substitute:

From the first part of this exercise we know that the probability that a student obtains a score greater than points on the exam is .

Now we will use the conditional probability formula:

We substitute:

The probability that a student who obtained a score greater than actually obtained a score greater than is .

We locate in our table the parameter corresponding to probability 0.2, (20%), which is :

So, if , then:

Now we locate in the table the parameter for probability , which is , which means:

So, if , then:

Low knowledge: up to points.
Acceptable knowledge: between and .
Excellent knowledge: from points onward.

1. Determine the percentage of the population that would obtain a coefficient between and .

Substituting values in the formula:

The percentage of the population that will obtain a score between and is .

2. What interval centered on contains of the population?

Since we want to take the from the center of the population, we take the interval that is between and .

We locate in the table the parameter for probabilities and :

We substitute and solve:

and

Therefore, the interval is: .

The centered interval that contains of the population will obtain a score between and .

3. In a population of individuals, how many individuals are expected to have a coefficient greater than ?

We substitute values in the formula, calculate the parameter, and locate the probability in the table:

Multiplying this probability by the individuals:

In a population of individuals, it is expected that of them have a coefficient greater than .

Where:

n: Number of families to choose.
p: Probability of selecting a family that has a telephone.
q: Complement of the probability.
To solve this type of exercise we will use the De Moivre-Laplace Theorem for Probability:

If is a binomial random variable with parameters and , , then can be approximated to a normal distribution with mean and standard deviation (where ) if the following two conditions are met:

Condition 1:
Condition 2:
Then, the binomial variable would be approximated by the normal variable .

Since , condition 1 is met.

Therefore, condition 2 is met.

Then we use the formula .

We substitute the data:

Now we use the normal distribution formula:

We substitute, operate, and locate the probability value in our normal distribution table:

When selecting families at random, there is a probability of of having selected at least families with a telephone.

We use the De Moivre-Laplace Theorem for Probability:

We check the 2 conditions:

First condition:

Second condition:

Since both conditions are met, we will use the formula:

We substitute:

Now we will use :

When answering a multiple-choice test exam at random, there is a probability of of passing.

1. What is the probability that at least of the cited households have at least two televisions?

We use the De Moivre-Laplace Theorem for Probability, checking if the 2 conditions are met:

Since both conditions are met, we will use the formula .

We substitute:

Now we will use .

We substitute:

2. What is the probability that between and households have at least two televisions?

Using the formula , we will substitute the value of the mean and the standard deviation :

The probability that between and households have at least televisions is .