Question

X 20 21 22 23 24

F(x) 22 8 11 6 3
Let x be the ages of students in a class. Given the frequency distribution F(x)
above, determine the following probabilities:
(a) P(x<24)=?

Answer 1

Answer:To find P(x < 24):

Sum the frequencies for x values less than 24:

P(x < 24) = F(20) + F(21) + F(22) + F(23)

P(x < 24) = 22 + 8 + 11 + 6

P(x < 24) = 47

Calculate the total number of students by summing all the frequencies:

Total number of students = F(20) + F(21) + F(22) + F(23) + F(24)

Total number of students = 22 + 8 + 11 + 6 + 3

Total number of students = 50

Now, divide the sum of the frequencies for x < 24 by the total number of students to find the probability:

P(x < 24) = 47 / 50

To express the probability as a decimal, divide 47 by 50:

P(x < 24) = 0.94

So, the probability that a randomly selected student from the class is younger than 24 years old is 0.94 or 94%.

Answer 2

`\text{P}(x < 24)=(47)/(50)`

`\text{P}(x \leq 23)=(47)/(50)`

`\text{P}(20 \leq x < 22)=(3)/(5)`

Explanation:

Given frequency distribution F(x):

`\begin{array}c\cline{1-6}x&20&21&22&23&24\\\cline{1-6}F(x)&22&8&11&6&3\\\cline{1-6}\end{array}`

where x is the ages of students in a class.

To find the given probabilities, sum the frequencies for all values of x that satisfy the given conditions and then divide by the total frequencies.

`\hrulefill`

To find the probability that x is less than 24, sum the frequencies for all values of x that are less than 24 and then divide by the total frequencies.

`\text{P}(x < 24)=(22+8+11+6)/(22+8+11+6+3)`

`\text{P}(x < 24)=(47)/(50)`

`\hrulefill`

To find the probability that x is less than or equal to 23, sum the frequencies for all values of x that are less than or equal to 23 and then divide by the total frequencies.

`\text{P}(x \leq 23)=(22+8+11+6)/(22+8+11+6+3)`

`\text{P}(x \leq 23)=(47)/(50)`

`\hrulefill`

To find the probability that x is greater than or equal to 20 but less than 22, sum the frequencies for values of x between 20 (inclusive) and 22 (exclusive) and then divide by the total frequencies.

`\text{P}(20 \leq x < 22)=(22+8)/(22+8+11+6+3)`

`\text{P}(20 \leq x < 22)=(30)/(50)`

`\text{P}(20 \leq x < 22)=(3)/(5)`