Question

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 the probability of obtaining a reading less than -2.74°C. Round your answer to 4 decimal places

Answer 1

We need to find the probability of obtaining a reading less than -2.74°C from a normal distribution with a mean of 0°C and a standard deviation of 1.00°C.

Using the standard normal distribution, we have:

z = (x - μ) / σ

where:

x = -2.74°C (the reading we want)

μ = 0°C (the mean)

σ = 1.00°C (the standard deviation)

Substituting the values, we get:

z = (-2.74 - 0) / 1.00 = -2.74

Using a standard normal distribution table or calculator, we find that the probability of obtaining a z-score less than -2.74 is approximately 0.0030.

Therefore, the probability of obtaining a reading less than -2.74°C from the batch of thermometers is approximately 0.0030.

Answer 2

Explanation:

We are given that the temperature readings are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. Let X be the temperature reading of a single thermometer selected at random. Then, X ~ N(0, 1).

We need to find the probability of obtaining a reading less than -2.74°C, which can be expressed mathematically as P(X < -2.74).

Using standard normal distribution tables or a calculator, we can find that the z-score corresponding to -2.74°C is:

z = (x - μ) / σ = (-2.74 - 0) / 1 = -2.74

The probability can be calculated as:

P(X < -2.74) = P(Z < -2.74) ≈ 0.0030 (rounded to 4 decimal places)

Therefore, the probability of obtaining a reading less than -2.74°C is approximately 0.0030.