Please help me I will give you the best answer

Suppose that x is a binomial random variable with n=5, p=. 3,and q=. 7.1. Write the binomial formula for this situation and list

Radda [10]

Answer:

$P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}$

Possible values of x: Any from 0 to 5.

$P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807$

$P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015$

$P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087$

$P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323$

$P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835$

$P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243$

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this question:

$n = 5, p = 0.3, q = 1 - p = 0.7$

So

$P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}$

Possible values of x: 5 trials, so any value from 0 to 5.

For each value of x calculate p(☓ =x)

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807$

$P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015$

$P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087$

$P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323$

$P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835$

$P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243$

8 0

3 years ago

Madeline invests $25,000 in an account that offers a compound interest rate of 9.5% per year. Which of the following is the corr

LekaFEV [45]

Use this equation: Amount after years=Initial investment*(1+Interest rate/time compounded yearly)^number of years*times compounded yearly

So A=25,000(1+.095/1)^8*1
Simplify
A=25000(1.095)^8
Simplify
A=25000(2.07)
Solve
A=$51,671.73

This equation can be used for all problems of this type.

5 0

2 years ago

Consider a binary code with 5 bits (0 or 1) in each code word. An example of a code word is 01001. In each code word, a bit is z

AleksandrR [38]

Answer:

the probability that a code word contains exactly one zero is 0.0064 (0.64%)

Step-by-step explanation:

Since each bit is independent from the others , then the random variable X= number of 0 s in the code word follows a binomial distribution, where

p(X)= n!/((n-x)!*x!*p^x*(1-p)^(n-x)

where

n= number of independent bits=5

x= number of 0 s

p= probability that a bit is 0 = 0.8

then for x=1

p(1) = n*p*(1-p)^(n-1) = 5*0.8*0.2^4 = 0.0064 (0.64%)

therefore the probability that a code word contains exactly one zero is 0.0064 (0.64%)

3 0

3 years ago

The distribution of passenger vehicle speeds traveling on a certain freeway in California is nearly normal with a mean of 73.7 m

Schach [20]

Answer:

A) 90.32%

B) 90.11%

C) 81.546 miles per hour

D)77.93%

Step-by-step explanation:

Given data:

speed = 73.7 miles/hour

standard deviation = 4.77 miles /hour

a) $P(X< 80) = P(z< \frac{80 - 73.7}{4.77}$

= P (z< 1.31)

= 0.90320 = 90.32%

b) $P(60 < X< 80) = P(\frac{60 - 73.7}{4.77} < z< \frac{80 - 73.7}{4.77})$

= P (-2.872< z< 1.31)

= P (z<1.31) - P(z<-2.87)

= 0.90320 - 0.00205

= 0.90115 = 90.11%

c) P(X <X) = 0.95

z = 1.645

$\frac{x - 73.7}{4.77} = 1.645$

x = 81.546 miles per hour

$d)P(X< 70) = P(z> \frac{70 - 73.7}{4.77})$

= P (z> -0.775)

= 0. 0.7793 = 77.93%

4 0

3 years ago

A fire truck has a long ladder that is made of 3 sections. Each section is 43/4 meters long.

Amanda [17]

You need to simplify the improper fraction into a mixed number.

43/4 = 10 3/4 (or 10.75)
You will want to have your answer as a mixed number, but multiplying might be easier if you use the decimal for calculations.

Now you multiply your mixed number or decimal by 3 (3 sections).

10.75 * 3 = 32.25 or 32 1/4

Your answer would be 32 1/4 meters.

Hope this helps :)

5 0

3 years ago