Mathematics: Please help Will give brainliest

Please help Will give brainliest - id11886972, jonmorton159

Ответ:

nikki8240

20.02.2020

40.13%

Step-by-step explanation:

We have been given that scores on the quantitative portion of an exam have a mean of 563 and a standard deviation of 144. Assume the scores are normally distributed.

We are asked to find the percentage of students taking the quantitative exam, who scored above 599.

First of all we will find z-score corresponding to 599 using z-score formula.

$z=\frac{x-\mu}{\sigma}$ , where,

z = z-score,

x = Random sample score,

$\mu$ = Mean,

$\sigma$ = Standard deviation.

$z=\frac{599-563}{144}$

$z=\frac{36}{144}$

z=0.25

Now we will use normal distribution table to find the probability of z-score greater than 0.25 that is P(z0.25) .

Using formula P(za)=1-P(z , we will get:

P(z0.25)=1-P(z

P(z0.25)=1-0.59871

P(z0.25)=0.40129

Let us convert 0.40129 into percentage as:

$0.40129\times 100\%=40.129\%\approx 40.13\%$

Therefore, approximately 40.13% of students taking the quantitative exam will score above 599.

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