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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 P9, the 9-percentile. This is the temperature reading separating the bottom 9% from the top 91%.. FInd P9 in °C.

User CloudWave
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Final answer:

To find the 9th percentile (P9) for a normal distribution with a mean of 0°C and a standard deviation of 1.00°C, we look for the z-score corresponding to the 9th percentile, which is approximately -1.34, and then use the formula X = Z × σ + μ to calculate P9 as -1.34°C.

Step-by-step explanation:

To find P9, the 9th-percentile of a normally distributed set of thermometer readings with a mean (average) of 0°C and a standard deviation of 1.00°C, we need to determine the temperature reading that separates the bottom 9% from the top 91%.

First, we find the z-score that corresponds to the 9th percentile of the standard normal distribution. This z-score represents the number of standard deviations below the mean.

Looking up the 9th percentile in the z-table, we find the z-score associated with P9, which is approximately -1.34.

Once the z-score has been identified, we use the z-score formula:

Z = (X - μ) / σ

We rearrange the formula to solve for X, which represents P9:

X = Z × σ + μ

Substituting our values, we get:

X = (-1.34 × 1.00°C) + 0°C = -1.34°C

The temperature reading that corresponds to P9 is -1.34°C.

User Henryk Borzymowski
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