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OverLord2011 [107]

2 years ago

11

Plz help me !!!!!!!!

1 answer:

Dovator [93]2 years ago

7 0

Answer: c) -1372

<u>Step-by-step explanation:</u>

$\sqrt[3]{2x} +6=-8\\\\\sqrt[3]{2x} =-14\\\\(\sqrt[3]{2x})^3 =(-14)^3\\\\2x=-2744\\\\\boxed{x=-1372}$

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Suppose a student's score on a standardize test to be a continuous random variable whose distribution follows the Normal curve.

blsea [12.9K]

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a) Percentage of students scored below 300 is 1.79%.

b) Score puts someone in the 90th percentile is 638.

Step-by-step explanation:

Given : Suppose a student's score on a standardize test to be a continuous random variable whose distribution follows the Normal curve.

(a) If the average test score is 510 with a standard deviation of 100 points.

To find : What percentage of students scored below 300 ?

Solution :

Mean $\mu=510$ ,

Standard deviation $\sigma=100$

Sample mean $x=300$

Percentage of students scored below 300 is given by,

$P(Z\leq \frac{x-\mu}{\sigma})\times 100$

$=P(Z\leq \frac{300-510}{100})\times 100$

$=P(Z\leq \frac{-210}{100})\times 100$

$=P(Z\leq-2.1)\times 100$

$=0.0179\times 100$

$=1.79\%$

Percentage of students scored below 300 is 1.79%.

(b) What score puts someone in the 90th percentile?

90th percentile is such that,

$P(x\leq t)=0.90$

Now, $P(\frac{x-\mu}{\sigma} < \frac{t-\mu}{\sigma})=0.90$

$P(Z< \frac{t-\mu}{\sigma})=0.90$

$\frac{t-\mu}{\sigma}=1.28$

$\frac{t-510}{100}=1.28$

$t-510=128$

$t=128+510$

$t=638$

Score puts someone in the 90th percentile is 638.

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