Assume that the readings at freezing on a bundle 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 the probability of obtaining a reading less than -1.344°C.
Find: P(Z < -1.344)
To find the probability P(Z < -1.344), we first need to standardize the value using the Z-score formula:
Z = (X - μ) / σ
Where X is the value for which we are finding the probability, μ is the mean, and σ is the standard deviation. Since the mean μ = 0°C and the standard deviation σ = 1.00°C, and we want to find the probability of obtaining a reading less than -1.344°C (X = -1.344), we can plug those values into the Z-score formula:
Z = (-1.344 - 0) / 1.00
Z = -1.344
We see that the Z-score is -1.344, which means the standardized value is already given. Now, we use a Z-table or a statistical calculator to find the probability P(Z < -1.344). Since Z-tables typically give the probability from the mean to the value, and our value is to the left of the mean (negative), we actually look up the probability for the positive Z-score and subtract it from 0.5 (the probability to the left of the mean in a standard normal distribution) to get the probability for the negative Z-score.
Therefore:
P(Z < -1.344) = 0.5 - P(0 < Z < 1.344)
Look up 1.344 in the Z-table to find P(0 < Z < 1.344). The value corresponding to a Z-score of 1.34 in the Z-table is 0.4102, and if we interpolate slightly for the additional 0.004, it would add a very small amount, which we might neglect for simplicity (otherwise, with interpolation or a statistical calculator, you might get a more precise value, but it would be very close to 0.4102). Thus:
P(Z < -1.344) ≈ 0.5 - 0.4102
P(Z < -1.344) ≈ 0.0898
So, the probability of obtaining a thermometer reading less than -1.344°C is approximately 0.0898 or 8.98%.