A population of values has a normal distribution with \mu=99.6 and \sigma=35.1.

Isa Trevino 2020-12-07 Answered

A population of values has a normal distribution with μ=99.6 and σ=35.1. You intend to draw a random sample of size n=84.
Find the probability that a single randomly selected value is between 98.5 and 100.7.
P(98.5<X<100.7)=?
Write your answers as numbers accurate to 4 decimal places.

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Expert Answer

Delorenzoz
Answered 2020-12-08 Author has 91 answers

Step 1
From the provided information,
Mean (μ)=99.6
Standard deviation (σ)=35.1
XN(99.6,35.1)
Sample size (n)=84
Step 2
The required probability that a single randomly selected value is between 98.5 and 100.7 can be obtained as:
P(98.5<X<100.7)=P(98.599.635.1<xμσ<100.799.635.1)
=P(0.031<Z<0.031)
=P(Z<0.031)P(Z<0.031)
=P(Z<0.031)[1P(z<0.031)]
=2P(Z<0.031)1
=2(0.5124)1=0.0248 (Using standard normal table)
Thus, the required probability is 0.0248.

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Jeffrey Jordon
Answered 2021-11-17 Author has 2070 answers

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