Assume that the readings at freezing on a bundle of thermometers are normally distributed with a mean of $0^{\circ} \mathrm{C}$ and a standard deviation of $1.00^{\circ} \mathrm{C}$. A single thermometer is randomly selected and tested. Find the probability of obtaining a reading tess than $-1.876^{\circ} \mathrm{C}$.
\[
P(Z< -1.876)=3.04
\]

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Step 1 ：Assume that the readings at freezing on a bundle of thermometers are normally distributed with a mean of \(0^{\circ} \mathrm{C}\) and a standard deviation of \(1.00^{\circ} \mathrm{C}\). A single thermometer is randomly selected and tested. The question is asking for the probability of obtaining a reading less than \(-1.876^{\circ} \mathrm{C}\).

Step 2 ：Since the readings are normally distributed, we can use the standard normal distribution (Z-distribution) to find this probability. The Z-score is calculated as \((X - \mu) / \sigma\), where X is the value we're interested in, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. In this case, X = -1.876, \(\mu\) = 0, and \(\sigma\) = 1. So, the Z-score is \((-1.876 - 0) / 1 = -1.876\).

Step 3 ：We want to find \(P(Z < -1.876)\), which is the area to the left of -1.876 under the standard normal curve.

Step 4 ：The probability of obtaining a reading less than \(-1.876^{\circ} \mathrm{C}\) is approximately 0.0303. This means that there is a 3.03% chance of randomly selecting a thermometer that gives a reading less than \(-1.876^{\circ} \mathrm{C}\).

Step 5 ：Final Answer: The final answer is \(\boxed{0.0303}\).