Question

The following numbers represent the highest temperatures (in Fahrenheit) during the last 18 days of a city. 68, 75, 87 ,99, 72, 72, 65, 64, 61, 70, 77 ,70 ,67, 73, 83, 55, 58, 54 a. Calculate the mean, variance, and the standard deviation of the above data. b. Calculate the first quartile, the third quartile, and the 90th percentile. Interpret these values.

Answer 1

a) The mean μ = 70.
`\overline 5`

The variance = 127.3203

The standard deviation is 11.28363

b) i) The first quartile is 63.25

ii) The third quartile is 75.5

iii) The 90th percentile is 88.2

Explanation:

a) The city temperatures ae presented as follows;

Fahrenheit; 68, 75, 87, 99, 72, 72, 65, 64, 61, 70, 77, 70, 67, 73, 83, 55, 58, 54

The data points in the sample, N = 18

The mean, μ = ∑
`x_i`/N

Where;

`x_i` = The value of each data point

N = The sample size = 18

From Microsoft Excel, we have;

∑
`x_i` = 1,270

∴ μ = 1,270/18 = 70.
`\overline 5`

The mean μ is 70.
`\overline 5`

The variance is given as follows;

`\sigma^2 =(\sum \left (x_i-\mu \right )^(2) )/(N - 1)}`

The variance = 127.3203

The standard deviation is given as follows;

`\sigma =\sqrt{(\sum \left (x_i-\mu \right )^(2) )/(N - 1)}`

We get σ = 11.28363

The standard deviation, σ = 11.28363

b) i) The first quartile, Q₁, is the (n + 1)/4th term

∴ The first quartile is the (18 + 1)/4 = 19/4 = 4.75th term

54, 55, 58, 61, 64, 65, 67, 68, 70, 70, 72, 72, 73, 75, 77, 83, 87, 99

The first quartile = 61 + (64 - 61)×0.75 = 63.25

ii) The third quartile, Q₃, is the 3×(n + 1)/4th term

∴ The third quartile is the 3×(18 + 1)/4 = 57/4 = 14.25th term

Therefore;

Q₃ = 75 + (77 - 75)×0.25 = 75.5

The third quartile is 75.5

iii) The 90th percentile = 90/100 × (18 + 1)th term = 17.1th term

∴ The 90th percentile = 87 + (99 - 87)×0.1 = 88.2