Question

Variable: Temperature
Count: 130 people
Mean: 36.80
Standard Deviation: 0.40
Min: 35.70
Q1: 36.55
Med: 36.85
Q3: 37.05
Max: 38.20

(a) Mr. Withey’s temperature is at the 20th percentile of the distribution. Interpret this value in context.
(b) What was Mr. Withey’s temperature?
(c) Find the mean temperature in degrees Fahrenheit. What does this suggest about the value of the “normal” body temperature?
(d) Calculate the standard deviation of the temperature readings in degrees Fahrenheit.

Answer 1

• The 20th percentile represent the temperature value which is less than or equal to the 20% of the temperature values in the distribution.
• The 20th percentile temperature is 36.5°C which is Mr. Whitney's temperature.
• The mean temperature value in °Fahrenheit is 98.249°F
• The standard deviation value of temperature in °F is 32.734°F

The 20th percentile represents the score at which less than or equal to 20% of the scores in a distribution may be found. This is the score at which less than or equal to 20% of the temperature values in the distribution may be found.

Mr. Whitney's temperature :

20th percentile ;

20% of (n)

Where n = count of temperature values = 130

20% × 130 = 26th term

The 26th falls in 36.5 temperature value.

The mean temperature reading in degree Fahrenheit :

°F = (9/5)°C + 32

Mean in °C = 36.805

°F = (9/5) × 36.805 + 32

°F = 66.249 + 32

= 98.249°F

The standard deviation of temperature reading in degree Fahrenheit :

°F = (9/5)°C + 32

Standard deviation in °C = 0.408

°F = (9/5) × 0.408 + 32

°F = 0.7344 + 32

= 32.7344°F

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