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inysia [295]
3 years ago
5

I need help please explain

Mathematics
1 answer:
Nitella [24]3 years ago
4 0
Where's the picture or the papper or the problem I would help but I can't see it
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Compute​ P(x) using the binomial probability formula. Then determine whether the normal distribution can be used to estimate thi
stich3 [128]

Answer:

The answers to the questions are;

A) P(X) where x = 12 is equal to 3.55 ×10⁻²

B) P(x) using the normal distribution is given by the probability distribution function and is equal to 3.453 ×10⁻²

C) The calculated probabilities of P(x=12) using the binomial probability formula and the probability distribution function differ by 9.7×10⁻⁴

D) The normal distribution be used to approximate this probability because     a. because np(1-p) %u2265 ⇒ np(1-p) ≥ 10

E) c. the value of fi represents the expected proportion of observation less than or equal to the ith data value.

Step-by-step explanation:

To solve the question, we note that

n = 58

p = 0.3

x = 12

Here we have n·p = 17.4 and n·q = 40.6 both ≥ 5 that is the normal distribution can be used to estimate the probability

A) We are to use the binomial probability formula to find P(X)

Where x = 12, we have P(12) = ₅₈C₁₂×0.3¹²×0.7⁴⁶

= 891794789340×0.000000531441×7.49×10⁻⁸ = 3.55 ×10⁻²

B) Using the standard distribution table we have

the z score for x = 12 given as z = \frac{x-\mu}{\sigma} = -1.55

From the normal distribution table, we have the probability that value is below 12 = .06057 while the normal probability distribution function which gives the probability of a number being 12 is given by

\frac{1}{\sigma\sqrt{2 \pi } } e^{\frac{-(x-\mu)^2}{2\sigma^2} }

where:

σ = Standard deviation = \sqrt{npq} = \sqrt{np(1-p)} = \sqrt{58*0.3*(1-0.3)} = 3.48999

μ = Sample mean = n·p = 58×0.3 =17.4

Therefore the probability density function is \frac{1}{3.49\sqrt{2 \pi } } e^{\frac{-(12-17.4)^2}{2*3.49^2} } = 3.453*10^{-2}

C)  The probabilities differ by 3.55 ×10⁻² -  3.453 ×10⁻² =  9.7×10⁻⁴

D) The normal distribution be used to approximate this probability because     n·p and n·p·(n-p) ≥ 5

E)   f_i represents the area under the curve towards left of the ith data observed in a normally distributed population.

Therefore, the value of fi represents the expected proportion of observation less than or equal to the ith data value  

5 0
3 years ago
There are two 5th grade classes at Jones Elementary School. Mrs. Roberts has 17 girls and 18 boys in her class. Mr. Simpson has
boyakko [2]

Answer:

The chance of a girl being picked is an 18/35 chance.

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
4x+3y=20<br> 2x+y=7<br> How do I solve this? I forgot how to do it :/ also no answer choice
Crank

So for this, I will be using the substitution method. Firstly, subtract both sides in the second equation by 2x: 4x+3y=20\\ y=7-2x

Next, substitute y in the first equation with 7-2x, then solve for x from there:

4x+3(7-2x)=20\\ 4x+21-6x=20\\ -2x+21=20\\ -2x=-1\\ x=\frac{1}{2}

Now that we got x, substitute it into either equation to solve for y:

4*\frac{1}{2}+3y=20\\ 2+3y=20\\ 3y=18\\ y=6\\ \\ 2*\frac{1}{2}+y=7\\ 1+y=7\\ y=6

In short, x = 1/2 and y = 6.

6 0
3 years ago
Question 22 Noella wraps cube-shaped boxes at her gift shop. The cost of wrapping a box depends on its side length. A box with a
AlekseyPX

Answer:

D B E

Step-by-step explanation:

8 0
3 years ago
Solve x2 + 1/2x +1/16=4/9
kolezko [41]

<u>Answer:</u>

x = 0.417 or x = -0.917

<u>Step-by-step explanation:</u>

We are given the following expression and we are to solve it for the variable x:

x ^2 + \frac { 1 } { 2 } x + \frac { 1 } { 1 6 } = \frac { 4 } { 9 }

We will find the least common multiple of 2. 6 and 9:

x^2 \times 144 +\frac{1}{2}x \times 144 +\frac{1}{16} \times 144 = \frac{4}{9} \times 144

Simplifying it to get:

144x^2+72x+9=64

144x^2+72x+9-64=0

144x^2+72x-55=0

Using the quadratic formula to solve for x:

x=\frac{-b \pm\sqrt{b^2-4ac} }{2a}

x=\frac{-72 \pm\sqrt{72^2-4(144)(-55)} }{2a}

x=\frac{5}{12} or 0.417

x=-\frac{11}{12} or -0.917

3 0
3 years ago
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