Mathematics, 29.07.2020 01:01 recon12759

Consider a binomial experiment with 15 trials and probability 0.35 of success on a single trial. (a) Use the binomial distribution to find the probability of exactly 10 successes (Round your answer to three decimal places.)
(b) Use the normal distribution to approximate the probability of exactly 10 successes. (Round your answer to three decimal places.) (
(c) Compare the results of parts (a) and (b).
A. These results are fairly different.
B. These results are almost exactly the same.

Answers: 2

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Answer from: quinteroalex2001

P(X = 10 ) = 0.0096

P(X = 10 ) = 0.0085

Option A is correct

Step-by-step explanation:

From the question we are told that

The sample size is n = 15

The probability of success is p = 0.35

The number of success we are considering is r = 10

Now the probability of failure is mathematically evaluated as

q = 1- p

substituting value

q = 1- 0.35

q = 0.65

Now using the binomial distribution to find the probability of exactly 10 successes we have that

$P(X = r ) = [\left n } \atop {r}} \right. ] * p^r * q^{n- r}$

substituting values

$P(X = 10 ) = [\left 15 } \atop {10}} \right. ] * p^{10}* q^{15- 10}$

Where $[\left 15 } \atop {10}} \right. ]$ mean 15 combination 10 which is evaluated with a calculator to obtain

$[\left 15 } \atop {10}} \right. ] = 3003$

$P(X = 10 ) = 3003 * 0.35 ^{10}* 0.65^{15- 10}$

P(X = 10 ) = 0.0096

Now using the normal distribution to approximate the probability of exactly 10 successes, we have that

P(X = r ) = P( r < X < r )

Applying continuity correction

P(X = r ) = P( r -0.5 < X < r +0.5)

substituting values

P(X = 10) = P( 10-0.5 < X < 10+0.5)

P(X = 10 ) = P( 9.5 < X < 10.5)

Standardizing

$P(X = r ) = P( \frac{9.5 - \mu }{\sigma } < \frac{X - \mu }{\sigma } < \frac{10.5 - \mu}{\sigma } )$

The where $\mu$ is the mean which is mathematically represented as

$\mu = n * p$

substituting values

$\mu = 15 * 0.35$

$\mu = 5.25$

The standard deviation is evaluated as

$\sigma = \sqrt{n * p * q }$

substituting values

$\sigma = \sqrt{15 * 0.35 * 0.65 }$

$\sigma = 1.8473$

Thus

$P(X = 10 ) = P( \frac{9.5 - 5.25 }{1.8473 } < \frac{X - 5.25 }{1.8473 } < \frac{10.5 - 5.25}{1.8473 } )$

P(X = 10 ) = P( 2.30 < Z < 2.842 )

P(X = 10 ) = P(Z < 2.842 ) - P(Z < 2.30 )

From the normal distribution table we obtain the P(Z < 2.841) as

P(Z < 2.841) = 0.99775

And the P(Z < 2.30)

P(Z < 2.30) = 0.98928

There value can also be obtained from a probability of z calculator at (Calculator dot net website)

P(X = 10) = 0.99775 - 0.98928

P(X = 10 ) = 0.0085

Looking at the calculated values for question a and b we see that the values are fairly different.

Answer from: Quest

its a trick question

step-by-step explanation:

dogs love water, just turn on the hose and it will run in circles chasing the water all day : )

Answer from: Quest

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf =

pf = 6,25 u

ver más en -

step-by-step explanation:

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