Assume the average BMI for adults in France is 24.5. Suppose the BMI for adults in...

Assume the average BMI for adults in France is 24.5. Suppose the BMI for adults in France is normally distributed with standard deviation σ=1.3σ=1.3.
1.(5) Find the probability that the BMI of one randomly selected adult is greater than 25.
2.(5) Find the probability that the BMI of one randomly selected adult is between 24 and 25, inclusive.
3.(5) Suppose the 15 adults from France are selected at random what is the probability that the sample mean of BMI is between 24 and 25, inclusive?

Solutions

Expert Solution

Solution :

P(x > 25) = 1 - P(x < 25)

= 1 - P[(x - ) / < (25 - 24.5) / 1.3)

= 1 - P(z < 0.3846)

= 1 - 0.6497

= 0.3503

Probability = 0.3503

P(24 < x < 25) = P[(24 - 24.5) / 1.3) < (x - ) / < (25 - 24.5) / 1.3) ]

= P(-0.3846 < z < 0.3846)

= P(z < 0.3846) - P(z < -0.3846)

= 0.6497 - 0.3503

= 0.2994

Probability = 0.2994

= / n = 1.3 / 15 = 0.3357

= P[(24 - 24.5) / 0.3357 < ( - ) / < (25 - 24.5) / 0.3357)]

= P(-1.4894 < Z < 1.4894)

= P(Z < 1.4894) - P(Z < -1.4894)

= 0.9318 - 0.0682

= 0.8636

Probability = 0.8636

orchestra answered 1 year ago

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