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bulgar [2K]
3 years ago
10

On the last math test in Kenzie’s class, 17 out of 25 students scored an 80 or higher. Which percent is closest to the experimen

tal probability that a student selected at random will score an 80 or higher on the next test?
Mathematics
1 answer:
kaheart [24]3 years ago
7 0

Answer

There is a 68 percent that a student selected a random will score an 80 or higher on the next test.

Explanation

You just need to find the percentage of 17 out of 25, or 17/25.

You can do this by dividing 17 by 25.

17/25 equals 0.68, or 68 percent.

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A pyramid with a square base of length 3cm and height 16cm
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6 0
3 years ago
The sugar content of the syrup in canned peaches is normally distributed. A random sample of n = 10 cans yields a sample standar
Serggg [28]

Answer:

The 95% confidence interval is given by:

3.30<σ<8.76

Step-by-step explanation:

1) Data given and notation

s=4.798 represent the sample standard deviation

\bar x represent the sample mean

n=10 the sample size

Confidence=95% or 0.95

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:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

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

df=n-1=10-1=9

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a tabel to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.025,9)" "=CHISQ.INV(0.975,9)". so for this case the critical values are:

\chi^2_{\alpha/2}=19.022

\chi^2_{1- \alpha/2}=2.700

And replacing into the formula for the interval we got:

\frac{(9)(4.798)^2}{19.022} \leq \sigma \frac{(9)(4.798)^2}{2.700}

10.892 \leq \sigma^2 \leq 76.736

Now we just take square root on both sides of the interval and we got:

3.30 \leq \sigma \leq 8.76

So the best option would be:

3.30<σ<8.76

7 0
3 years ago
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