A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

theopenstudyowl

  • one year ago

@

  • This Question is Closed
  1. pooja195
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @?

  2. theopenstudyowl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Michele_Laino

  3. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    question #1 I think the first option, namely: \[\Large \begin{gathered} {\mu _{\bar x}} = p \hfill \\ {\sigma _{\bar x}} = \frac{\sigma }{{\sqrt n }} \hfill \\ {\mu _{\bar p}} = \bar x \hfill \\ {\sigma _{\bar p}} = \sqrt {\frac{{p\left( {1 - p} \right)}}{n}} \hfill \\ \end{gathered} \]

  4. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    sorry, please wait, I didn't see the other options

  5. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    First question: I think the fifth option, namely the last option

  6. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    question #8 when I flip three coins one time, I have probability p= 1/8 to get three heads probability q= 7/8 to get other events Now the requested probability is equal to the product p*q, namely p*q= (1/8)*(7/8)=...

  7. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Question #9 the mean is 75 and the standard deviation is: \[\sigma = \sqrt {100 \times \frac{{75}}{{100}} \times \frac{{25}}{{100}}} = 4.33\] now those value referred to the population, becomes: mean=75/100=... standard deviation = 4.33/100=...

  8. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Question #7 here the standardized variable t is: t=(6.8-6)/3.2= 0.25 Now the probability to get a length less than 6 inches, is 40.13%, here I'm referring to the "erf" function table, so the probability to get a length greater than 6 inches is 100-40.13= 59.87 so we have 1000*59.87% = 598.7 crabs, whose shell has a length greater than 6 inches. Then the requested proportion is: 50/598.7=...

  9. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Question #4 here we have to apply the binomial distribution, so we can write: \[\Large 0.0148 = \left( {\begin{array}{*{20}{c}} n \\ {15} \end{array}} \right)\frac{1}{{{2^n}}}\] that equation is checked for n=20

  10. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    question #5 here we can compute these two parameters: mean=400*0.78=312 standard deviation = sqrt(400*0.78*0.22)= 8.28 standardized varaible t= (312-300)/8.28= 1.5 so, using the table of the "erf" function, we get: probability= 0.5-0.4332=...

  11. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Question #2 I think it is the last option, namely normalcdf(-E99,50,400,7.14) since we have this: mean value = 0.85*400=... standard deviation = sqrt(400*0.85*0.15)= 7.14

  12. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Question #3 the probability to get a double is 1/12, whereas the probability to get not a double is 11/12 so,I think that the requested probability is: \[\Large P = {\left( {\frac{1}{{12}}} \right)^4}{\left( {\frac{{11}}{{12}}} \right)^4}\]

  13. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Question #1 your answer is right!

  14. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Question #6 we have: minimum value= 500*0.4=200 probability of success= 0.35 number of tests=500 so it is the last option: using the Central Limit Thorem- binomcdf(500, 0.35, 200)

  15. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.