In a large lecture class, the professor announced that the scores on a recent exam were normally distributed with a range from 51 to 87. Using the Empirical Rule μ ± 3σ to estimate σ, how many students would you need to sample to estimate the true mean score for the class with 90 percent confidence and an error of ±2?

The standard deviation of the scores on a recent exam is 6.

The sample size required is 25.

Step-by-step explanation:

Let X = scores of students on a recent exam.

It is provided that the random variable X is normally distributed.

According to the Empirical rule, 99.7% of the normal distribution is contained in the range, μ ± 3σ.

That is, P (μ - 3σ < X < μ + 3σ) = 0.997.

It is provided that the scores on a recent exam were normally distributed with a range from 51 to 87.

This implies that:

P (51 < X < 87) = 0.997

So,

μ - 3σ = 51...(i)

μ + 3σ = 87...(ii)

Subtract (i) and (ii) to compute the value of σ as follows:

  μ -     3σ =    51

(-)μ + (-)3σ = (-)87

______________

-6σ = -36

σ = 6

Thus, the standard deviation of the scores on a recent exam is 6.

The (1 - α)% confidence interval for population mean is given by:

The margin of error of this interval is:

Given:

MOE = 2

σ = 6

Confidence level = 90%

Compute the z-score for 90% confidence level as follows:

*Use a z-table.

Compute the sample required as follows:

Thus, the sample size required is 25.

step-by-step explanation:

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

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step-by-step explanation: