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White raven [17]
2 years ago
15

The following table shows the number of hours some teachers in two schools expect students to spend on homework each week: Schoo

l A 9 14 15 17 17 7 15 6 6 School B 12 8 13 11 19 15 16 5 8 Part A: Create a five-number summary and calculate the interquartile range for the two sets of data. (6 points) Part B: Are the box plots symmetric? Justify your answer. (4 points)
Mathematics
1 answer:
Lerok [7]2 years ago
4 0

Answer:

For school A: Minimum=6, Q₁=6.5, Median= 14, Q₃=16, Maximum=17, IQR=9.5

For school B: Minimum=5, Q₁=8, Median= 12, Q₃=15.5, Maximum=19, IQR=7.5

No, the box plots are not symmetric.

Step-by-step explanation:

Part A

The given data sets are

School A : 9,14,15,17,17,7,15,6,6

School B : 12,8,13,11,19,15,16,5,8

Arrange the data in ascending order.

School A : 6,6,7,9,14,15,15,17,17

School B : 5,8,8,11,12,13,15,16,19

Divide each data set in four equal parts.

School A : (6,6),(7,9),14,(15,15),(17,17)

School B : (5,8),(8,11),12,(13,15),(16,19)

For school A:

Minimum=6, Q₁=6.5, Median= 14, Q₃=16, Maximum=17

Interquartile range of the data is

IQR=Q_3-Q_1=16-6.5=9.5

For school B:

Minimum=5, Q₁=8, Median= 12, Q₃=15.5, Maximum=19

Interquartile range of the data is

IQR=Q_3-Q_1=15.5-8=7.5

Part B:

The box plots are not symmetric because the data values are different. Five number summary and IQR of both the data set are different.

Guest
1 year ago
what are the outliers
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We want to estimate the total weight of grain on the field based on the data on a simple random sample of 5 piles out of 200. The population and sample sizes are N=200 & n=5 respectively.

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2) To calculate a bound on the error of estimates, we need to find the sample standard deviation.

The sample standard deviation is

 S=\sqrt{\frac{1}{5-1}\sum_{i=1}^{5}(y_i-\overline{y})^2}\: \: \: =0.9486

Then, the standard error of \widehat{Y} is

\sigma_{\widehat{Y}} =\sqrt{\frac{N^2S^2}{n}\bigg(\frac{N-n}{N}\bigg)}\: \: \:=83.785

Hence, a 95% bound on the error of estimates is

[\pm z_{0.025}\times \sigma_{\widehat{Y}}]\: \: \: =[\pm 1.96\times 83.875]\: \: \: =[\pm 164.395]

3) Let x_1,x_2,...,x_5 denotes the total weight of the sampled piles.

Mean total weight of the sampled piles is

\overline{x}=\frac{1}{5}\sum_{i=1}^{5}x_i=45

The sample ratio is

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