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The partially completed box plot in Figure 1 shows the distribution of daily mean air temperatures using the data from the large data set for Beijing in 2015 - Edexcel - A-Level Maths Statistics - Question 2 - 2019 - Paper 1

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The partially completed box plot in Figure 1 shows the distribution of daily mean air temperatures using the data from the large data set for Beijing in 2015. An ou... show full transcript

Worked Solution & Example Answer:The partially completed box plot in Figure 1 shows the distribution of daily mean air temperatures using the data from the large data set for Beijing in 2015 - Edexcel - A-Level Maths Statistics - Question 2 - 2019 - Paper 1

Step 1

Complete the box plot in Figure 1 showing clearly any outliers.

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Answer

To complete the box plot, we must first calculate the IQR. From the provided values:

  1. Find Q1Q_1 and Q3Q_3:

    • The three lowest values are 7.6°C, 8.1°C, and 9.1°C, thus:
    • Q1Q_1 can be taken as the median of the lower half of data, which will be a value greater than 7.6°C but less than 8.1°C.
    • For Q3Q_3, determined similarly, will be a value greater than 9.1°C.
  2. Calculate IQR:

    • IQR = Q3Q1Q_3 - Q_1.
  3. Identify Outliers:

    • Lower Limit = Q11.5imesIQRQ_1 - 1.5 imes IQR.
    • Upper Limit = Q3+1.5imesIQRQ_3 + 1.5 imes IQR.
    • Any data point outside of these limits is considered an outlier.
  4. Plotting:

    • The box will be drawn based on the quartiles, whiskers drawn to the nearest values within limits, and any outliers will be marked distinctly.

Step 2

Using your knowledge of the large data set, suggest from which month the two outliers are likely to have come.

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Answer

The two outliers are likely to come from October as it typically has the coldest temperatures between May and October in Beijing. Therefore, extreme cold temperatures are expected in this month.

Step 3

Show that, to 3 significant figures, the standard deviation is 5.19°C.

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Answer

To calculate the standard deviation, we use the formula:

S=(xixˉ)2n1S = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}

Given: Ss=4952.906S_s = 4952.906, therefore:

s=Ssn=4952.90618426.8s = \frac{S_s}{n} = \frac{4952.906}{184} \approx 26.8

Consequently, standard deviation simplifies to: 26.8184=5.19\sqrt{\frac{26.8}{184}} = 5.19 to 3 significant figures.

Step 4

Using Simon's model, calculate the 10th to 90th interpercentile range.

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Answer

The 10th and 90th percentiles for a normally distributed variable can be found using the Z-table:

  • For the 10th percentile, Z0.101.2816Z_{0.10} \approx -1.2816.
  • For the 90th percentile, Z0.901.2816Z_{0.90} \approx 1.2816.

Using the formula: P=μ+ZσP = \mu + Z \cdot \sigma

  1. 10th Percentile Calculation:

    • P10=22.6+(1.28165.19)18.3P_{10} = 22.6 + (-1.2816 \cdot 5.19) \approx 18.3.
  2. 90th Percentile Calculation:

    • P90=22.6+(1.28165.19)29.9P_{90} = 22.6 + (1.2816 \cdot 5.19) \approx 29.9.
  3. Interpercentile Range:

    • Range = P90P10=29.918.311.6P_{90} - P_{10} = 29.9 - 18.3 \approx 11.6.

Step 5

State two variables from the large data set for Beijing that are not suitable to be modeled by a normal distribution. Give a reason for each answer.

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Answer

  1. Rainfall: Rainfall data is typically not normally distributed as it is often skewed (not symmetric), with many days having zero rain, leading to a left skew.

  2. Wind Speed: Wind speed data often follows a different statistical distribution (like Rayleigh distribution) since it usually has a lower bound of zero and does not exhibit symmetry, resulting in a distribution that is not normal.

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