A population of values has a normal distribution with \mu=116.3 and \sigma=27.5. You intend to draw a random sample of size n=249. Find the probability that a single randomly selected value is greater than 117.3. P(X > 117.3) =? Write your answers as numbers accurate to 4 decimal places.

facas9 2020-10-23 Answered
A population of values has a normal distribution with μ=116.3 and σ=27.5. You intend to draw a random sample of size n=249.
Find the probability that a single randomly selected value is greater than 117.3.
P(X>117.3)=?
Write your answers as numbers accurate to 4 decimal places.
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Expert Answer

Daphne Broadhurst
Answered 2020-10-24 Author has 109 answers
Step 1
Suppose, X follows Normal distribution with mean μ=116.3 and standard deviation σ=27.5.
The z-score for x is defined as,
Z=xμσ
=x116.327.5
Step 2
The probability of a single randomly selected value is greater than 117.3
P[X>117.3]
=P[X116.327.5>117.3116.327.5]
=P[Z>127.5]
=1P[Z127.5]
=1ϕ(0.036)
=10.5144=0.4856
The value of ϕ(0.036) is taken from the Normal distribution table.
Therefore, the probability of a single randomly selected value is greater than 117.3 is 0.4855.
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