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Complete Question
Wild deer population weight has standard deviation of 50 lb. What is the probability of existence of a random sample (size 100) weight within ± 5 lb. of population mean weight?
Answer:
0.68269
Explanation:
We solve using z score formula
z = (x-μ)/σ/√n where
x is the raw score
μ is the population mean
σ is the population standard deviation.
From the question:
Standard deviation = 50 lb.
Sample size = 100
The weight is within ± 5Ib
For +5 lb
Z =+ 5/50/ √100
Z = +5/50/10
Z = +5/5
Z = 1
Using the Z table to find the probability of the Z score.
P(Z = 1) = 0.84134
For - 5 lb
Z = - 5/50/ √100
Z = - 5/50/10
Z = - 5/5
Z = -1
Using the Z table to find the probability of the Z score.
P(Z = -1 ) = 0.15866
Hence,
P (-Z<x<Z)
P(Z = +1 ) - P(Z = -1)
0.84134 - 0.15866
= 0.68269
The probability of existence of a random sample (size 100) weight within ± 5 lb. of population mean weight is 0.68269