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
LiRa [457]
2 years ago
9

businessText message users receive or send an average of 62.7 text messages per day. How many text messages does a text message

user receive or send per hour? (2 pts) What is the probability that a text message user receives or sends three messages per hour? (3 pts) What is the probability that a text message user receives or sends more than three messages per hour? (3 pts)
Mathematics
1 answer:
KiRa [710]2 years ago
6 0

Answer:

(a) The probability that a text message user receives or sends three messages per hour is 0.2180.

(b) The probability that a text message user receives or sends more than three messages per hour is 0.2667.

Step-by-step explanation:

Let <em>X</em> = number of text messages receive or send in an hour.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em>.

It is provided that users receive or send 62.7 text messages in 24 hours.

Then the average number of text messages received or sent in an hour is: \lambda=\frac{62.7}{24}= 2.6125.

The probability of a random variable can be computed using the formula:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0, 1, 2, 3, ...

(a)

Compute the probability that a text message user receives or sends three messages per hour as follows:

P(X=3)=\frac{e^{-2.6125}(2.6125)^{3}}{3!} =0.21798\approx0.2180

Thus, the probability that a text message user receives or sends three messages per hour is 0.2180.

(b)

Compute the probability that a text message user receives or sends more than three messages per hour as follows:

P (X > 3) = 1 - P (X ≤ 3)

              = 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

             =1-\frac{e^{-2.6125}(2.6125)^{0}}{0!}-\frac{e^{-2.6125}(2.6125)^{1}}{1!}-\frac{e^{-2.6125}(2.6125)^{2}}{2!}-\frac{e^{-2.6125}(2.6125)^{3}}{3!}\\=1-0.0734-0.1916-0.2503-0.2180\\=0.2667

Thus, the probability that a text message user receives or sends more than three messages per hour is 0.2667.

You might be interested in
50 point question for this geometry test answer
aniked [119]

Answer:

d

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Help Please! I will give brainliest!!
seropon [69]
The answer is 32.7 b so the answer is 105 over 10
6 0
2 years ago
Read 2 more answers
Can anyone help me ASAP!
slega [8]

Answer:

B

Step-by-step explanation:

6 0
2 years ago
PLS HELP OLL GIVE BRAINLIEST
Rudiy27

Answer:

yes they are the same

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
The age of a father is 3 times the age of his son. 5 years ago, the father was 4
LUCKY_DIMON [66]

Answer:

Step-by-step explanation:

x = father's age

y = son's age

Nowadays:

3x = y

5 years ago:

4(x - 5) = y - 5

4x - 20 = y - 5

4x = y + 15

I hope I've helped you.

3 0
3 years ago
Other questions:
  • The chapter is 8 pages long. Kosta read 1/4 of the chapter aloud. Then Christina read three pages to the class. How many pages h
    7·2 answers
  • Help please!!!!!!!!!!!
    12·1 answer
  • <img src="https://tex.z-dn.net/?f=%20%7B7%7D%5E%7B%20-%202%7D%20%20%7B%28%20-%203%29%7D%5E%7B2%7D%20" id="TexFormula1" title=" {
    11·2 answers
  • Which number line shows the sum of -8, 4, and -2?
    14·1 answer
  • Help me pls i dont understand this
    13·1 answer
  • Helppp, 60 points!!!
    9·2 answers
  • Just help me again for 2 times
    15·1 answer
  • The endpoints of EF¯¯¯¯¯¯¯¯ are E(−4,12) and F(3,15). Find the coordinates of the midpoint M..
    14·1 answer
  • Please help with my homework. I will give a brainlist out. Only solve 28 please.
    10·2 answers
  • An aircraft on a reconnaissance mission takes off from its home base and flies 550 miles at a bearing of s 46° e to a location i
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!