In: Statistics and Probability
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?
Solution :
1.
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
2.
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
3.
= / 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