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antoniya [11.8K]
3 years ago
6

In a city school, 70% of students have blue eyes, 45% have dark hair, and 30% have blue eyes and dark hair. What is the probabil

ity (rounded to the nearest whole percent) that a randomly selected student will have dark hair, given that the student has blue eyes?
Hint:
P(A|B)=P(A∩B) / P(B)


46%


43%


21%


67%
Mathematics
1 answer:
Travka [436]3 years ago
4 0

<u>Answer:</u>

The correct answer option is 43%.

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

We are given that 70% of students have blue eyes, 45% have dark hair, and 30% have blue eyes and dark hair.

We are to find the probability of a student getting selected will have dark hair with blue eyes.

P(D|B) = P(D∩B) / P(B)

Substituting the given values in it to get:

P(D|B) = 0.3 / 0.7 = 0.428

Rounding it to the nearest whole percent we get:

0.428 × 100 = 42.8% ≈ 43%

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Step-by-step explanation:

1) Data given and notation

s^2 =16 represent the sample variance

s=4 represent the sample standard deviation

\bar x represent the sample mean

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A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

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The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

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Since the Confidence is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.01,19)" "=CHISQ.INV(0.99,19)". so for this case the critical values are:

\chi^2_{\alpha/2}=36.191

\chi^2_{1- \alpha/2}=7.633

And replacing into the formula for the interval we got:

\frac{(19)(16)}{36.191} \leq \sigma^2 \leq \frac{(19)(16)}{7.633}

8.400 \leq \sigma^2 \leq 39.827

So the 98% confidence interval for the variance in the pounds of impurities would be 8.400 \leq \sigma^2 \leq 39.827.

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