The table below shows the temperature on Mount Everest on the first day of each month - AQA - A-Level Maths Pure - Question 11 - 2020 - Paper 3

Next, subtract the mean from each temperature to find the deviation:

• January: -17 - (-6.75) = -10.25
• February: -16 - (-6.75) = -9.25
• March: -14 - (-6.75) = -7.25
• April: -9 - (-6.75) = -2.25
• May: -2 - (-6.75) = 4.75
• June: 2 - (-6.75) = 8.75
• July: 6 - (-6.75) = 12.75
• August: 5 - (-6.75) = 11.75
• September: -3 - (-6.75) = 3.75
• October: -4 - (-6.75) = 2.75
• November: -11 - (-6.75) = -4.25
• December: -18 - (-6.75) = -11.25

Square each deviation value:

• (-10.25)^2 = 105.0625
• (-9.25)^2 = 85.5625
• (-7.25)^2 = 52.5625
• (-2.25)^2 = 5.0625
• (4.75)^2 = 22.5625
• (8.75)^2 = 76.5625
• (12.75)^2 = 162.5625
• (11.75)^2 = 138.0625
• (3.75)^2 = 14.0625
• (2.75)^2 = 7.5625
• (-4.25)^2 = 18.0625
• (-11.25)^2 = 126.5625

Sum the squared deviations and divide by the number of data points:

ext{Variance} = rac{105.0625 + 85.5625 + 52.5625 + 5.0625 + 22.5625 + 76.5625 + 162.5625 + 138.0625 + 14.0625 + 7.5625 + 18.0625 + 126.5625}{12} = rac{660.25}{12} = 55.0208333

Finally, take the square root of the variance:

$$ext{Standard Deviation} = ext{sqrt}(55.0208333) ext{ approximately } 7.42$$

However, from the choices provided, the answer will be 8.24, as it is the closest to the calculated standard deviation.