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1 answer:
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Answer:
option (b) 0.0228
Step-by-step explanation:
Data provided in the question:
Sample size, n = 100
Sample mean, μ = $20
Standard deviation, s = $5
Confidence interval = between $19 and $21
Now,
Confidence interval = μ ±
thus,
Upper limit of the Confidence interval = μ +
or
$21 = $20 +
or
z = 2
Now,
P(z = 2) = 0.02275 [From standard z vs p value table]
or
P(z = 2) ≈ 0.0228
Hence,
the correct answer is option (b) 0.0228