In: Statistics and Probability
A population is normally distributed with
muμequals=100100
and
sigmaσequals=2525.
a. nbspa. |
Find the probability that a value randomly selected from this
population will have a value greater than
115115. |
b. |
Find the probability that a value randomly selected from this
population will have a value less than
9090. |
c. |
Find the probability that a value randomly selected from this
population will have a value between
9090 and115115. |
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a. P(xgreater than>115115)equals=
(Round to four decimal places as needed.)b. P(xless than<9090)equals=
(Round to four decimal places as needed.)c. P(9090less than<xless than<115115)equals=
(Round to four decimal places as needed.)
Answer: A population is normally distributed with
μ =100 and
σ = 25.
a) Find the probability that a value randomly selected from this population will have a value greater than 115.
P(X > 115) = P(x - μ/σ > 115 - 100 / 25)
P(X > 115) = P(Z > 0.60)
P(X > 115) = 0.2743
Therefore, the probability that a value randomly selected from this population have a value greater than 115 would be 0.2743.
b) Find the probability that a value randomly selected from this population will have a value less than 90.
P(X < 90) = P(x - μ/σ < 90 - 100 / 25)
P(X < 90) = P(Z < - 0.40)
P(X < 90) = 0.3446
Therefore, the probability that a value randomly selected from this population have a value less than 115 would be 0.3446.
c) Find the probability that a value randomly selected from this population will have a value between 90 and 115.
P(90 X < 115) = P(90 - 100 /25 < Z < 115 - 100 /25)
P(90 X < 115) = P(- 0.40 < Z < 0.60)
P(90 X < 115) = P(Z< 0.60) - P(Z< -0.40)
P(90 X < 115) = 0.7258 - 0.3446
P(90 X < 115) = 0.3812
Therefore, the probability that a value randomly selected from this population have a value between 90 and 115 would be 0.3812.