Question

A distribution of scores is approximately normal with a mean of 78 and a standard deviation of 8.6. Which of the following equations can be used to find the score x above which 33 percent of the scores fall?
(A) 0.44=1.78
(B) 0.67= 78
(C) 0.33= -78
(D) 0.44=1.78
(E) 0.67-8.68

Answer 1

Answer:

The correct answer is (D) $0.44=\frac{x-78}{8.6}$

Step-by-step explanation:

If we check the value in the Normal Distribution table, the x score that falls above 33% is given by:

P(z>x)=0.33

1 - P(z≤x)=0.33

1 - 0.33=P(z≤x)

0.67=P(z≤x)⇒ z=0.44

Also

z=(x-μ)/σ

where:

μ: mean of the distribution

σ: standard deviation of the distribution

Hence,

0.44=(x-78)/8.6

Answer 2

The correct option is D).

Step-by-step explanation:

Given :

Mean, $\mu = 78$

Standard Deviation, $\sigma = 8.6$

Solution :

In the Normal Distribution table, the z score that falls above 33% is

$\rm P(z>x)=0.33$

$\rm 1-P(z\leq x) = 0.33$

$\rm P(z\leq x) = 1-0.33=0.67$

z = 0.44

We know that z score is given by

$z = \dfrac{x-\mu}{\sigma}$

$0.44=\dfrac{x-78}{8.6}$

The correct option is D).

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