Question

@

Answer 1

@?

Answer 2

@Michele_Laino

Answer 3

Bazinga

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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)=...

Answer 8

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=...

Answer 9

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=...

Answer 10

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

Answer 11

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=...

Answer 12

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

Answer 13

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

Answer 14

Question #1
your answer is right!

Answer 15

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)