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FrozenT [24]
3 years ago
7

Surveys indicate that 5% of the students who took the SATs had enrolled in an SAT prep course. 30% of the SAT prep students were

admitted to their first choice college, as were 20% of the other students. You overhear a classmate say he got into the college he wanted. What is the probability he didn't take an SAT prep course?
What steps would I take to solve this problem?
Mathematics
1 answer:
Nesterboy [21]3 years ago
4 0

Answer: The probability he didn't take an SAT prep course = 0.985

Step-by-step explanation:

Let us first assume that he took SAT prep.

Surveys indicate that 5% of the students who took the SATs had enrolled in an SAT prep course. 30% of the SAT prep students were admitted to their first choice college. That is,

30/ 100 of 5 = 0.3 × 5 = 1.5

The probability he did take an SAT prep course and got admission into the college of first choice will be

P(prep) = 1.5 / 100 = 0.015

The probability he didn't take an SAT prep course will be:

P(not prep) = 1 - P(prep)

P(not prep) = 1 - 0.015

P(not prep) = 0.985

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Answer:

A 90% confidence interval to estimate the mean DRP score in Henrico County Schools  is =32±2.7280

i.e. C.I.  [29.3,34.7]

Hence the results are not significant.

Step-by-step explanation:

Given:

Total no of students =44

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And all 4 students score also given.

To Find:

90 % confidence interval and conclusion on it.

Solution:

Here for C.I. we required mean, standard deviation and no of students.

So mean =32,S.D.=11 and n=44

Therefore , For 90 % interval Zscore is Z=1.645

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=32±1.645[11/Sqrt(44)]

=32±1.645[1.6583]

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Now Here to commit on calculated interval we should check  value of Z-score

So we need to calculate the sample mean.

Sample mean =Sum of all values /Total no of values.

=[40+ 26 +39+ 14+ 42+ 18 +25+ 43+ 46 +27 +19+ 47+ 19 +26+ 35 +34+ 15+ 44 40+ 38+ 31+ 46 +52 +25+ 35 +35+ 33 +29 +34 +41 +49+ 28+ 42+ 47 +35 +48 +22 33+ 41 +51 +27 +14+ 54 +45]/44

=34.86

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P(Z≥1.72465)=0.08544

For 0.05 significance level  the results are not significant at  p<0.01.

Researcher claimed mean is not correct upto 0.05 significant level

i.e calculated mean and Sample mean are different.

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6 0
3 years ago
If a triangle has sides that are 21 and 6 what is the range for third side x?
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Step-by-step explanation:

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Then

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15<x<27

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