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Assume that the readings at freezing on a batch of thermometers are normally distributed with a...

Assume that the readings at freezing on a batch of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. Find P72, the 72-percentile. This is the temperature reading separating the bottom 72% from the top 28%. P72 = °C

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Solution :

Given that,  

mean = = 0

standard deviation = = 1.00

Using standard normal table ,

P(Z < z) = 72%

P(Z < 0.58) = 0.72

z = 0.58

Using z-score formula,

x = z * +

x = 0.58 * 1.00 + 0 = 0.58

P72 = 0.580 C

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