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zmey [24]
3 years ago
9

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, what percentage of students scored below 300? Enter as a percentage to the nearest tenth of a percent. % (b) What score puts someone in the 90th percentile? Start by finding z such that P(Z < z) = 0.90, then find what test score has that z value. Round to the nearest whole score:
Mathematics
1 answer:
blsea [12.9K]3 years ago
5 0

Answer:

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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