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pantera1 [17]
4 years ago
10

Helppp 100pts

Mathematics
1 answer:
frosja888 [35]4 years ago
8 0

Answer:

.83

Step-by-step explanation:

If you take 100 shows 60 shows got favorable response and in that 50 shows were successful.

So probability for a show to be successful if it got a favorable response is = 50/60 = 0.83

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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
Which of the following are solutions to the inequality below? Select all that apply.
AysviL [449]
Answer: j=8 and j=11

Explanation: 7(8) and 7(11) is equal to 56 and 77. 56 and 77 are both greater than 30. Therefore, the answer is j=8 and j=11.
5 0
3 years ago
If KLMN is a square, then ___________________.
beks73 [17]
I do believe it's A since a rhombus is a shape with all 4 sides the same length m8
4 0
3 years ago
Read 2 more answers
(8-9)-(9+3)*5<br> 5+(5+5+6)+9<br> (5/3-3)<br> 2*3+(4*2)/2
pashok25 [27]

Answer:

If you need any further help, feel free to let me know

Step-by-step explanation:

The answers are

(8-9)-(9+3)*5=-61

5+(5+5+6)+9=30

(5/3-3)= -4 over 3 or -4/3 or -1.3 or -1 1/3 depends on what your choices are but all those are correct

2*3+(4*2)/2=10

6 0
3 years ago
HELPPPP PLEASEEEEEEEREEREEEE
Maksim231197 [3]

Answer:

the correct answer is b :)))))!!!

7 0
3 years ago
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