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 kg.

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Complete Question

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.

88.36%

Step-by-step 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%

probability for j is 1/4

while

a is 1/6

is there any photos or whatever?