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According to one study, brain weights of men are normally distributed with a mean of 1.40 kg and a standard deviation of 0.11 kg.
Determine the percentage of all samples of three men that have mean brain weights within 0.1 kg of the population mean brain weight of 1.40 kg.
Answer:
88.36%
Explanation:
When given random samples, the formula for z score that we use is:
z = (x-μ)/σ/√n
where x is the raw score
μ is the population mean
σ is the population
n is number of random samples
mean = 1.40 kg
Standard deviation = 0.11 kg
n = 3
x = mean brain weights within 0.1 kg of the population mean brain weight of 1.40 kg.
= 0.1kg ± 1.40kg
Hence,
For 1.50kg
z = 1.50 - 1.40/0.11/√3
z = -1.57
Probability value from Z-Table:
P(x = 1.50) = 0.9418
For 1.30kg
z = 1.30 - 1.40/0.11/√3
= -1.57
Probability value from Z-Table
P(x= 1.30) = 0.0582
P(x = 1.50) - P(x= 1.30)
0.9418 - 0.0582
= 0.8836
Converting to percentage,
0.8836 × 100
= 88.36%