Download 3.4 The Normal Distribution Probability for continuous random

3.4 The Normal Distribution
Probability for continuous random variables.
Example 1. Let X be the value where the pointer of the given spinner lands.
0
0.75
0.25
0.5
Notice 0 ≤ X < 1 and P (0 ≤ X < 1) = 1. Find the follwoing:
a) P (X = 0.9)
b) P (0 ≤ X < 1/4)
c) P (0.75 ≤ X ≤ 0.9)
Continuous random variables cannot be represented using histograms or tables, so we have to find
another way to understand continuous random variables. Note: If X is a continuous random variable,
P (X = k) = 0. (Notice that there are infinitely many points to spread the probabilities over.)
Revisit the histogram. The area of each bar give the probability of the random variable X. Since the
width of the bar is 1, the height also gives the value of probability. For continuous random variables,
the area of a certain point is zero and the height in this case does not gives the probability. We call the
function of the hight as probability density function (PDF). Thus the area under this density function
gives us the probability.
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Note: P (a ≤ X ≤ b) = area under the probability density function’s curve between the values x = a
and x = b.
Normal Distribution: A common continuous probability distribution.
• Many natural phenomena obey a normal distribution (or close to it)
Examples: weights of blue birds, heights of men in Chile, wait times at a hospital, the amount of
solution in an IV bag, the pounds of rabbit food in a 50-pound bag.
• A normally distributed random variable has the property that it can take any value in the interval
(−∞, ∞). (Thus, it is a continuous random variable.) We can view probabilities associated with
a continuous random variable that is normally distributed as areas under the density function.
P (a ≤ X ≤ b) =area under the normal curve on the interval [a, b]. The area under the entire
curve is 1.[P (S) = 1.]
The random variable X has a normal distribution with the mean µ and standard deviation σ on the
interval (−∞, ∞) if the probability P (a ≤ X ≤ b) that X is between a and b is the area under the
standard normal curve given by
(x−µ)2
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y = √ e− 2σ2
σ 2π
on the interval [a, b], where π ≈ 3.14159 and e ≈ 2.71828.
Properties of the Normal curve:
(1) It is bell shaped..
(2) It is symmetric about x = µ.
(3) It lies above the x-axis (positive value).
(4) It approaches, but is never equal to, 0 on both the positive and negative x-axis.
(5) The curve “bends downward” on the interval (µ−σ, µ+σ) and “bends upward” outside this interval.
(6) The area under the entire curve is exactly 1.
Note: The mean of the standard normal distribution is µ = 0 and the standard deviation σ = 1.
Convention: The random variable associated with the standard normal distribution is designated by
Z.
Finding Probabilities of Normal Distributions Using Your Calculator:
(1) Press DISTR (2nd → VARS).
(2) Select normalcdf( (Option 2).
(3) Enter the lower bound of the interval, comma key, upper bound of the interval, comma, the mean,
comma, the standard deviation, and close with the parentheses.
(4) Press ENTER.
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Example 2. Let Z be a random variable that has the standard normal distribution. Find the following.
(Round your answer to 4 decimal places.)
a) P (−1 ≤ Z ≤ 1)
b) P (2.31 ≤ Z ≤ 2.31)
c) P (0.75 < Z ≤ 1.17)
d) P (Z < 2.31)
e) P (Z ≥ 2.31)
Example 3. Assume X is distributed normally with mean µ = 70 and standard deviation σ = 5.25.
Find the following. (Round your answer to 4 decimal places.)
a) P (64.75 < X < 75.25)
b) P (55 < X < 78)
c) P (X ≤ 66.54)
d) P (X > 72.30)
Finding the Value a so that P (X ≤ a) = k
(1) Press DISTR (2nd → VARS).
(2) Select invNorm( (Option 3).
(3) Enter the area below the value a (i.e., the known value), comma, the mean, comma, the standard
deviation, and the closed parentheses.
(4) Press ENTER.
Example 4. Assume Z has the standard normal distribution.
a) Find a so that P (Z ≤ a) = 0.6754. (Round your answer to 4 decimal places.)
b) Find a so that P (Z ≥ a) = 0.6754. (Round your answer to 4 decimal places.)
c) Find a so that P (a ≤ Z ≤ a) = 0.6754. (Round your answer to 4 decimal places.)
Example 5. Assume that X is a random variable that is normally distributed with mean µ = 70 and
standard deviation σ = 5.25.
a) Find a so that P (Z ≤ a) = 0.6754. (Round your answer to 4 decimal places.)
b) Find a so that P (Z ≥ a) = 0.6754. (Round your answer to 4 decimal places.)
c) Find a and b symmetric about the mean so that P (a ≤ Z ≤ b) = 0.6754. (Round your answer to 4
decimal places.)
Example 6. According to data for a certain city, the weekly earnings of workers are normally distributed
with a mean of $700 and a standard deviation of $60.
a) What is the probability that a worker selected at random from the city makes more than $805?
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b) What minimum weekly earnings would put you in the top 15% of wage earners? Round your answer
to the nearest cent.
c) What symmetric interval of wages about the mean comprises 64% of wage earners? Round your
answers to the nearest cent.
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