1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
sattari [20]
3 years ago
7

Suppose there is a 17.5 % probability that a randomly selected person aged 35 years or older is a smoker. In​ addition, there

is a 25.5 % probability that a randomly selected person aged 35 years or older is female comma given that he or she smokes.
What is the probability that a randomly selected person aged 35 years or older is female and smokes?
Would it be unusual to randomly select a person aged 35 years or older who is female and smokes ?
Mathematics
2 answers:
nexus9112 [7]3 years ago
8 0

Answer:

The answer is 43% or 0.43

Step-by-step explanation:

Given:

Probability that person selected aged 35years or older is a smoker = 17.5%

Probability that person selected aged 35years or older is a female comma =25.5%

Solution:

a) Probability (35yrs or older is a smoker) + probability (35years or older is a female comma)

= 17.5% + 25.5%

= 43% 0r 0.43

b) No, it would not be unusual because it is by chance, it may and it may not.

Sophie [7]3 years ago
3 0

Answer:

a)  P(U and T) = 0.045

b) YES!, a randomly selected person aged 35 years and older is a female and smokes are an unusual.

Step-by-step explanation:

Let T represents the event that the randomly selected person is a smoker

Let U represents the event that the randomly selected person is female.

Given that; there is a 17.5% probability that a randomly selected person agef 35 years or older is a smoker.

i.e P(T) = 0.175

and there is a 25.5% probability that a randomly selected person aged 35 years or older is a female, given that he or she smokes.

i.e P (U|T) = 0.255

Then;

P (U|T) = \frac{P(U and T)}{P(T)}

0.255 = \frac{P(U and T)}{0.175}

P(U and T) = 0,255 × 0.175

P(U and T) = 0.044625

P(U and T) = 0.045

∴ The probability that a randomly selected person aged 35 years or older is female and smokes is 0.045

b)

The probability that a randomly selected person aged 35 years or older is female and smokes is 0.045.

It is seen that the probability is less than 0.05; thus illustrating that the randomly selected person aged 35 years or older is female and smokes is an unusual event.

∴ YES!, a randomly selected person aged 35 years and older is a female and smokes are an unusual.

You might be interested in
Find the product 5.39 divided by 0.31. (please show your work and how you got the answer)​
n200080 [17]

Answer:

\frac{5.39}{0.31} X100= \frac{539}{31}≈17.388 (expand this equation by multiplying both numerator and denominator by 100

Step-by-step explanation:

7 0
2 years ago
Please help me answer this question. I have a quiz tomorrow :(​
babymother [125]

Answer:

Y E S

Step-by-step explanation:

6 0
4 years ago
A rectangle’s area and perimeter have the same numerical value. The length of the rectangle is 10 feet.
KiRa [710]

Answer:

The width of the rectangle is 2.5 feet

Step-by-step explanation:

Here, we want to calculate the width of the rectangle given its length

Let the area = perimeter = x

Mathematically;

area of rectangle = L * B = 10B

The perimeter of the rectangle = 2(L + B)

Also;

x = 10B

Let’s equate this to the perimeter

x = 10B = 2(L + B)

10B = 2(10 + B)

10B = 20 + 2B

10B -2B = 20

8B = 20

B = 20/8

B = 5/2

B = 2.5

3 0
4 years ago
Suppose n people, n ≥ 3, play "odd person out" to decide who will buy the next round of refreshments. The n people each flip a f
blondinia [14]

Answer:

Assume that all the coins involved here are fair coins.

a) Probability of finding the "odd" person in one round: \displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1}.

b) Probability of finding the "odd" person in the kth round: \displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1} \cdot \left( 1 - n \cdot \left(\frac{1}{2}\right)^{n - 1}\right)^{k - 1}.

c) Expected number of rounds: \displaystyle \frac{2^{n - 1}}{n}.

Step-by-step explanation:

<h3>a)</h3>

To decide the "odd" person, either of the following must happen:

  • There are (n - 1) heads and 1 tail, or
  • There are 1 head and (n - 1) tails.

Assume that the coins here all are all fair. In other words, each has a 50\,\% chance of landing on the head and a

The binomial distribution can model the outcome of n coin-tosses. The chance of getting x heads out of

  • The chance of getting (n - 1) heads (and consequently, 1 tail) would be \displaystyle {n \choose n - 1}\cdot \left(\frac{1}{2}\right)^{n - 1} \cdot \left(\frac{1}{2}\right)^{n - (n - 1)} = n\cdot \left(\frac{1}{2}\right)^n.
  • The chance of getting 1 heads (and consequently, (n - 1) tails) would be \displaystyle {n \choose 1}\cdot \left(\frac{1}{2}\right)^{1} \cdot \left(\frac{1}{2}\right)^{n - 1} = n\cdot \left(\frac{1}{2}\right)^n.

These two events are mutually-exclusive. \displaystyle n\cdot \left(\frac{1}{2}\right)^n + n\cdot \left(\frac{1}{2}\right)^n  = 2\,n \cdot \left(\frac{1}{2}\right)^n = n \cdot \left(\frac{1}{2}\right)^{n - 1} would be the chance that either of them will occur. That's the same as the chance of determining the "odd" person in one round.

<h3>b)</h3>

Since the coins here are all fair, the chance of determining the "odd" person would be \displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1} in all rounds.

When the chance p of getting a success in each round is the same, the geometric distribution would give the probability of getting the first success (that is, to find the "odd" person) in the kth round: (1 - p)^{k - 1} \cdot p. That's the same as the probability of getting one success after (k - 1) unsuccessful attempts.

In this case, \displaystyle p = n \cdot \left(\frac{1}{2}\right)^{n - 1}. Therefore, the probability of succeeding on round k round would be

\displaystyle \underbrace{\left(1 - n \cdot \left(\frac{1}{2}\right)^{n - 1}\right)^{k - 1}}_{(1 - p)^{k - 1}} \cdot \underbrace{n \cdot \left(\frac{1}{2}\right)^{n - 1}}_{p}.

<h3>c)</h3>

Let p is the chance of success on each round in a geometric distribution. The expected value of that distribution would be \displaystyle \frac{1}{p}.

In this case, since \displaystyle p = n \cdot \left(\frac{1}{2}\right)^{n - 1}, the expected value would be \displaystyle \frac{1}{p} = \frac{1}{\displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1}}= \frac{2^{n - 1}}{n}.

7 0
3 years ago
Solve for x under the assumption that x &gt; 0. Enter your answer in interval notation using grouping symbols.
Delicious77 [7]
<span>Starting with x−24/x<10 multiply everything by x. x^2-24<10x Subtract 10x from both sides. x^2-10x-24 < 0 Factor. (x-12)(x+2) < 0 (x+2) is always positive since x > 0, but (x-12) is negative when x is a value between 0 and 12. This negative value for (x-12) satisfies that the left side is less than 0. Thus the solution set for x is (0,12).</span>
3 0
3 years ago
Other questions:
  • Mario was given some birthday money. He puts the money in an account. Every month after that, he deposits the same amount of mon
    9·2 answers
  • Find the value of x and the value of y.
    14·1 answer
  • Please help with my homework. It should be easy
    15·1 answer
  • Only four people can serve on a committee. how many committees can be selected from a group of 6 people
    15·1 answer
  • Explain whether the function g(x) = x^9 + x^5 is borh an even function and an odd function.
    9·1 answer
  • Plz help me i need it
    8·2 answers
  • Solve 0.667a+5=1.667a
    15·2 answers
  • What is the slope of the line that passes through (10,-1) and (-4,7)
    11·1 answer
  • Drag each one to the correct set
    6·1 answer
  • Max built a rectangular Prism and a rectangular Pyramid using beach sand. The prism and the pyramid have the same base area and
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!