Mathematics, 22.04.2020 16:26 Emptypockets451

In the country of United States of Heightlandia, the height measurements of ten-year-old children are approximately normally distributed with a mean of 54.4 inches, and standard deviation of 7.9 inches.

A) What is the probability that a randomly chosen child has a height of less than 46.35 inches?

Answer=

(Round your answer to 3 decimal places.)

B) What is the probability that a randomly chosen child has a height of more than 38.6 inches?

Answer=

Answers: 3

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Answers

Answer from: rebeckas0102

Given Information:

Mean height of ten year children = μ = 54.4 inches

Standard deviation = σ = 7.9 inches

Required Information:

A) What is the probability that a randomly chosen child has a height of less than 46.35 inches?

B) What is the probability that a randomly chosen child has a height of more than 38.6 inches?

$P(X < 46.35) = 15.38\%$

$P(X 38.6) = 97.73\%$

Explanation:

Let X is the random variable that represents height measurements of ten-year-old children with mean of 54.4 inches and standard deviation of 7.9 inches.

A) The probability that a randomly chosen child has a height of less than 46.35 inches is given by

$P(X < 46.35) = P(Z < \frac{x - \mu}{\sigma} )$

$P(X < 46.35) = P(Z < \frac{46.35 - 54.4}{7.9} )$

P(X < 46.35) = P(Z < -1.02)

From the z-table , the z-score corresponding to -1.02 is 0.1538

P(X < 46.35) = 0.1538

$P(X < 46.35) = 15.38\%$

Therefore, 15.38% children has a height of less than 46.35 inches.

B) The probability that a randomly chosen child has a height of more than 38.6 inches is given by

$P(X 38.6) = 1 - P(Z < \frac{x - \mu}{\sigma} )$

$P(X 38.6) = 1 - P(Z < \frac{38.6 - 54.4}{7.9} )$

P(X 38.6) = 1 -P(Z < -2)

From the z-table , the z-score corresponding to -2 is 0.0227

P(X 38.6) = 1 -0.0227

P(X 38.6) = 0.9773

$P(X 38.6) = 97.73\%$

Therefore, 97.73% children has a height of greater than 38.6 inches.

Answer from: Quest

answer: what in the world is this

step-by-step explanation:

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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