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FrozenT [24]
4 years ago
9

The table below shows data from a survey about the amount of time students spend doing homework each week. The students were eit

her in college or in high school:
High Low Q1 Q3 IQR Median Mean σ
College 50 5 7.5 15 7.5 11 13.8 6.4
HS 16 0 9.5 14.5 5 13 . 10.7 5.3
Which of the choices below best describes how to measure the spread of this data? (Hint: Use the minimum and maximum values to check for outliers.) (2 points) A:Both spreads are best described with the IQR.
B:Both spreads are best described with the standard deviation.
C:The college spread is best described by the IQR. The high school spread is best described by the standard deviation.
D:The college spread is best described by the standard deviation. The high school spread is best described by the IQR.
Mathematics
1 answer:
Nadusha1986 [10]4 years ago
4 0

Answer:

Option A is correct

Step-by-step explanation:

Given is a table which shows data from a survey about the amount of time students spend doing homework each week.

          College    HS

High    50            16

Low     5                0

Q1       7.5             9.5

Q3      15              14.5

 IQR    7.5             5

Median  11             13

Mean     13.8         10.7

Std dev 6.4            5.3

1.5IQR         11.25         7.5

Q1-1.5IQR     -9.38      -12

Q3+1.5IQR    37.5       29.5

WE find that there is only one outlier in college with 50 as high.

Except that all others are well within IQR range.  Hence outliers are minimum

So option A is right.

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7 0
3 years ago
construct a 90% confidence interval of the population proportion using the giver information x=74 n=150
FrozenT [24]

Answer:

The 90% confidence interval of the population proportion is (0.43, 0.56).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for population proportion <em>p</em> is:

CI=\hat p\pm z_{\alpha/2}\ \sqrt{\frac{\hat p(1-\hat p)}{n}}

The information provided is:

<em>X</em> = 74

<em>n</em> = 150

Confidence level = 90%

Compute the value of sample proportion as follows:

\hat p=\frac{X}{n}=\frac{74}{150}=0.493

Compute the critical value of <em>z</em> for 90% confidence level as follows:

z_{\alpha/2}=z_{0.10/2}=z_{0.05}=1.645

*Use a <em>z</em>-table.

Compute the 90% confidence interval of the population proportion as follows:

CI=\hat p\pm z_{\alpha/2}\ \sqrt{\frac{\hat p(1-\hat p)}{n}}

     =0.493\pm 1.645\times \sqrt{\frac{0.493(1-0.493)}{150}}\\\\=0.493\pm 0.0672\\\\=(0.4258,\ 0.5602)\\\\\approx (0.43,\ 0.56)

Thus, the 90% confidence interval of the population proportion is (0.43, 0.56).

3 0
3 years ago
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