Two-Way Chi-Square test Using Excel - Statistics Homework Help

Using this data from December at Rio Ruidoso to answer the following questions.

Dec
4.1
4.45
5.75
5.05
5.2
4.95
4.6
4.55
4.6
4.4
4.25
4.15
4.15
4.05
4.1
4.2
4.2
4.2
3.9
3.65
3.55
3.25
3.15
3.4
3.45
3.6
3.7
3.55
3.4
3.25
3.35

3. Assume that the data from December at Rio Ruidoso is normally distributed. Compute the following probabilities if one day is chosen at random.

a) The probability that the stream flow will be at least 5.00 ft3/sec.

b) The probability that the stream flow will be at most 5.91 ft3/sec.

c) The probability that the stream flow will be at least 3.08 ft3/sec.

d) The probability that the stream flow will be at most 2.95 ft3/sec.

e) The probability that the stream flow will be between 2.5 ft3/sec and 3.5 ft3/sec.

f) The probability that the stream flow will be between 4.00 ft3/sec and 5.56 ft3/sec.

Solution: From the data we obtain that

\[\bar{X}=4.069,\,\,\,s=0.634\]

The population standard deviation is not known, so we use s as an approximation

(a) \(\Pr \left( X\ge 5.00 \right)=\Pr \left( \frac{X-4.069}{0.634}\ge \frac{5.00-4.069}{0.634} \right)=\Pr \left( Z\ge 1.467 \right)=0.071188\)

(b) \(\Pr \left( X\le 5.91 \right)=\Pr \left( \frac{X-4.069}{0.634}\le \frac{5.91-4.069}{0.634} \right)=\Pr \left( Z\le 2.90 \right)=0.998\)

(c) \(\Pr \left( X\ge 3.08 \right)=\Pr \left( \frac{X-4.069}{0.634}\ge \frac{3.08-4.069}{0.634} \right)=\Pr \left( Z\ge -1.5596 \right)=0.94056\)

(d) \(\Pr \left( X\le 2.95 \right)=\Pr \left( \frac{X-4.069}{0.634}\le \frac{2.95-4.069}{0.634} \right)=\Pr \left( Z\le -1.76447 \right)=0.0388\)

(e) \(\Pr \left( 2.5\le X\le 3.5 \right)=\Pr \left( \frac{2.5-4.069}{0.634}\le \frac{X-4.069}{0.634}\le \frac{3.5-4.069}{0.634} \right) \) \[=\Pr \left( -2.47381\le Z\le -0.89749 \right)=0.178\]

(e) \(\Pr \left( 4\le X\le 5.6 \right)=\Pr \left( \frac{4-4.069}{0.634}\le \frac{X-4.069}{0.634}\le \frac{5.6-4.069}{0.634} \right) \) \[=\Pr \left( -0.10933\le Z\le 2.413 \right)=0.5356\]

4. Repeat question 3 (a-f) when 7 days are chosen at random.

Solution: Now we have

(a) \(\Pr \left( \bar{X}\ge 5.00 \right)=\Pr \left( \frac{\bar{X}-4.069}{0.634/\sqrt{7}}\ge \frac{5.00-4.069}{0.634/\sqrt{7}} \right)=\Pr \left( Z\ge 3.88 \right)=0.000052\)

(b) \(\Pr \left( \bar{X}\le 5.91 \right)=\Pr \left( \frac{\bar{X}-4.069}{0.634/\sqrt{7}}\le \frac{5.91-4.069}{0.634/\sqrt{7}} \right)=\Pr \left( Z\le 7.67 \right)=1.00000\)

(c) \(\Pr \left( \bar{X}\ge 3.08 \right)=\Pr \left( \frac{\bar{X}-4.069}{0.634/\sqrt{7}}\ge \frac{3.08-4.069}{0.634/\sqrt{7}} \right)=\Pr \left( Z\ge -4.12617 \right)=0.999982\)

(d) \(\Pr \left( \bar{X}\le 2.95 \right)=\Pr \left( \frac{\bar{X}-4.069}{0.634/\sqrt{7}}\le \frac{2.95-4.069}{0.634/\sqrt{7}} \right)=\Pr \left( Z\le -4.66834 \right)=0.00000152\)

(e) \(\Pr \left( 2.5\le \bar{X}\le 3.5 \right)=\Pr \left( \frac{2.5-4.069}{0.634/\sqrt{7}}\le \frac{\bar{X}-4.069}{0.634/\sqrt{7}}\le \frac{3.5-4.069}{0.634/\sqrt{7}} \right) \)

\[=\Pr \left( -6.5451\le Z\le -2.37453 \right)=0.008786\]

(e) \(\Pr \left( 4\le \bar{X}\le 5.6 \right)=\Pr \left( \frac{4-4.069}{0.634/\sqrt{7}}\le \frac{\bar{X}-4.069}{0.634/\sqrt{7}}\le \frac{5.6-4.069}{0.634/\sqrt{7}} \right)\) \[=\Pr \left( -0.28925\le Z\le 6.38 \right)=0.6138\]

5. Again, using the data from December and supposing that the data is approximately normally distributed, find the following.

a) The 30th percentile.

Solution: Assuming the distribution is normal with mean 4.069 and standard deviation 0.634, the 30^th percentile of a standard normal distribution is computed as

\[{{\Phi }^{-1}}\left( 0.30 \right)=-0.5244\]

which means that the 30^th percentile of the distribution is

\[{{P}_{30}}=4.069-0.5244\times 0.634=3.74\]

b) The score that separates the top 70 percent from the bottom 30 percent.

Solution: This corresponds to the 30^th percentile 3.74

c) P84

Solution: The 84^th percentile is computed as

\[{{P}_{84}}=4.069+0.634\times {{\Phi }^{-1}}\left( 0.84 \right)=4.7\]

6. Using the data from December, answer the following.

a) Approximately how many scores fall within 2 standard deviations from the mean?

Solution: We know that approximately 95% of the scores fall within 2 standard deviations. This means that approximately

\[0.95\times 31=29.45\approx 29\]

scores fall within 2 standard deviations.

b) At least how many scores fall within 1.94 standard deviations from the mean?

Solution: We have that

\[\Pr \left( -1.94\le Z\le 1.94 \right)=0.94762\]

which means that approximately

\[0.94762\times 31=29.37623\]

scores fall within 1.94 standard deviations, which means that at least 29 scores fall within 1.94 standard deviations

c) Approximately how many scores fall within 1 standard deviation from the mean?

Solution: We have that

\[\Pr \left( -1\le Z\le 1 \right)=0.683\]

which means that approximately

\[0.683\times 31=21.16337\approx 21\]

scores fall within 1 standard deviations

d) At least how many scores fall within 2.75 standard deviations from the mean?

Solution: We have that

\[\Pr \left( -1.94\le Z\le 1.94 \right)=0.94762\]

which means that approximately

\[0.994\times 31=30.81525\approx 31\]

scores fall within 2.75 standard deviations, which means that at least 30 scores fall within 2.75 standard deviations

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