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notsponge [240]
3 years ago
13

Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. If

75 women are randomly​ selected, find the probability that they have a mean height between 63 and 65 inches.
A. 0.9811
B. 0.0188
C. 0.2119
D. 0.3071

Mathematics
1 answer:
djyliett [7]3 years ago
5 0

Answer:

Option a. 0.9811

Step-by-step explanation:

First, you need to use a normal distribution table to solve this. If you don't have it, then see the attached table so you can guide yourself.

Now, we want to know the probability of randomly select 75 women whose height have a mean between 63 and 65 inches, knowing that in general, the mean is 63.5 and standard deviation of 2.5

To do this, you should calculate first the Z score value for both heights, one for 63 inches and the other with 65 inches. Then, with those values, we'll look the table to get the area under the curve, and thus, the probability. To calculate Z use the following expression:

Z = x - μ / (σ/√n)

Where:

μ: mean height

σ: standard deviation

x: height required

n: sample population

We have all the data so let's calculate both values of Z:

Z1 = 63 - 63.6 / (2.5/√75) = -2.08

Z2 = 65 - 63.6 / (2-5/√75) = 4.85

With these Z score, let's watch the table, and see which area belongs to,

In the case of Z1, the area is 0.0188, while Z2 is 1.

To know the probability, all we need to do is substract those values:

P = 1 - 0.0188 = 0.9812

Option a) is the most accurate and closest result to this.

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Answer:

20 disparos

Step-by-step explanation:

Primera etapa es de determinar cuantos tiros de Camilo acertaron al blanco.

Sabemos que en su última práctica, obtuvo 99 puntos con una mezcla de 5, 8 y 10 puntos.  Eso puedo se exprimir así:

5x + 8y + 10x = 99

Sabemos también que el obtuvo tantos tiros de 8 puntos que de 10 puntos, entonces

y = z

Podemos mezclar las 2 ecuaciones y substituir y por z:

5x + 8y + 10y = 99

5x +18y = 99

Una ecuación, dos variables... no es fácil... pero son números pequeños y se puede intentar soluciones.  Entonces, cuantas veces podemos multiplicar 5 y 18 para obtener 99?

El más simple es de hacer la tabla de multiplicación de 18 y ver cual número nos deja con un multiple de 5.

18 x 1 = 18 (99 - 18 = 81, no un multiple e 5)

18 x 2 = 36 (99 - 36 = 63, no un multiple de 5)

18 x 3 = 54 (99 - 54 = 45, SI, un multiple de 5)

18 x 4 = 72 (99 - 72 = 27, no un multiple de 5)

18 x 5 = 90 (99 - 90 = 9, no un multiple de 5)

Entonces, sabemos que y = 3 y z = 3

5x + 18 (3) = 99

5x + 54 = 99

5x = 45

x = 9

El tiró 9 veces por 5 puntos, 3 veces por 8 puntos y 3 veces pour 10 puntos.

En total tiró 15 veces.

Si acertó 75% (3/4) de las veces, cuantos tiros total?

\frac{15}{3/4} = \frac{15 * 4}{3} = 20

Camilo tiró 20 veces en total.

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