## theopenstudyowl one year ago @

1. pooja195

@?

2. theopenstudyowl

@Michele_Laino

3. theopenstudyowl

Bazinga

4. Michele_Laino

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

5. Michele_Laino

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

6. Michele_Laino

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

7. Michele_Laino

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)=...

8. Michele_Laino

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=...

9. Michele_Laino

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=...

10. Michele_Laino

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

11. Michele_Laino

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=...

12. Michele_Laino

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

13. Michele_Laino

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: $\Large P = {\left( {\frac{1}{{12}}} \right)^4}{\left( {\frac{{11}}{{12}}} \right)^4}$

14. Michele_Laino

15. Michele_Laino

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)

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