Mathematics, 11.07.2021 14:00 Michael321

14. In a statistics class with 15 males and 13 females, five students are selected to put problems on the board. What is the probability that: a. 3 females and 2 males are selected? b. all five students selected are males? c. all five students selected are females? d. at least one male is selected?

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Answer from: puchie1225

a) 0.3056 = 30.56% probability that 3 females and 2 males are selected.

b) 0.0306 = 3.06% probability that all five students selected are males.

c) 0.0131 = 1.31% probability that all five students selected are females.

d) 0.9869 = 98.69% probability that at least one male is selected.

Step-by-step explanation:

The students are chosen from the sample without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

In this question:

15 + 13 = 28 students, which means that N = 28

5 are selected, which means that n = 5

13 females, which means that k = 13

a. 3 females and 2 males are selected?

3 females, so this is P(X = 3).

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 3) = h(3,28,5,13) = \frac{C_{13,3}*C_{15,2}}{C_{28,5}} = 0.3056$

0.3056 = 30.56% probability that 3 females and 2 males are selected.

b.all five students selected are males?

0 females, so this is P(X = 0).

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,28,5,13) = \frac{C_{13,0}*C_{15,5}}{C_{28,5}} = 0.0306$

0.0306 = 3.06% probability that all five students selected are males.

c. all five students selected are females?

This is P(X = 5). So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 5) = h(5,28,5,13) = \frac{C_{13,5}*C_{15,0}}{C_{28,5}} = 0.0131$

0.0131 = 1.31% probability that all five students selected are females.

d.at least one male is selected?

Less than five females, so:

P(X < 5) = 1 - P(X = 5) = 1 - 0.0131 = 0.9869

0.9869 = 98.69% probability that at least one male is selected.

Answer from: Quest

How much is 32 out of 40 written as a percentage? convert fraction (ratio) 32 / 40 answer: 80%

Answer from: Quest

(a) common ratio in the given sequence is, 3

(b) $6.973568802 \times 10^9$ push ups will do kendall do on the 20th week

step-by-step explanation:

geometric sequence is a sequence of numbers that follows a pattern were the next term is found by multiplying by a constant called the common ratio(r)

(a)

given the sequence: 6 , 18 , 54 , 162 ,

this sequence is geometric sequence where each term is 3 times the previous term.

therefore,

common ratio(r) = 3.

(b)

a formula for nth term of geometric sequence is given by;

$a_n = a_1r^{n-1}$ , where r is the common ratio between the successive terms and $a_1$ is the first term,

to find how many push ups will kendall do on the 20th week.

here. n = 20th term.

in the given sequence first term $(a_1) = 6$

then;

$a_n = a_ 1 \cdot r^{n-1}$

$a_{20} = 6 \cdot 3^{20-1}$

simplify:

$a_{20} = 6.973568802 \times 10^9$

therefore, $6.973568802 \times 10^9$ push ups will do kendall do on the 20th week

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