The daily low temperature in Woodland, in December is normally distributed with mean 39.6°F and a standard deviation of 4.1°F. (a) Find the probability that the low temperature on a given day in Woodland, in December, will be less than 32°F. (b) What temperature separates the coldest 4% of all low temperatures in Woodland, in December, from the rest?
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
The daily low temperature in Woodland, in December is
deviation of 4.1°F.
(a) Find the probability that the low temperature on a given day in Woodland, in December, will be less than 32°F.
(b) What temperature separates the coldest 4% of all low temperatures in Woodland, in December, from the rest?
Introduction:
For a normally distributed random variable, X with mean μ, and standard deviation σ, the z-score is given as follows:
z = (x – μ)/σ.
(a)
Denote T as the low temperature on a randomly selected day in December in the given place. It is mentioned that T has a normal distribution with parameters, mean μ = 39.6 degrees Fahrenheit, and standard deviation σ = 4.1 degrees Fahrenheit.
The z-score for a day with T = 32 degrees Fahrenheit is:
z
= (T – μ)/σ
= (32 – 39.6) / 4.1
≈ –1.85.
The probability that the low temperature on a given day will be less than 32 degrees Fahrenheit is:
P (T < 32)
= P (Z < –1.85).
From the standard normal distribution table or using the Excel formula: =NORM.S.DIST(–1.85,TRUE), it is obtained that P (Z ≤ –1.85) = 0.3216 (correct to 4 decimal places).
Thus,
= 1 – 0.9868
= 0.0322.
Thus, the probability that the low temperature on a given day will be less than 32 degrees Fahrenheit is 0.0322.
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