Answer :
Final answer:
41.3% of the female student population weighs more than 67 kg. 20% of female students weigh less than 57.44 kg.
Explanation:
To solve this problem, we will use the concept of a 'Z-score' in statistics, which measures how many standard deviations an element is from the mean. First, calculate the Z score, then use a Z-table or normal distribution calculator for the probability.
a) To find the percentage that exceeds 67 kg, first, we calculate the Z score using the formula:
Z = (X - μ) / σ
Where:
X = observation (67 kg in this case)
μ = mean (65 kg)
σ = standard deviation (9 kg)
So, Z = (67 - 65) / 9 = 0.22
From standard normal tables, we know the probability for a Z score of 0.22 is .5871. This is the percentage below or equal to 67 kg. So, 100 - 58.7 = 41.3% would be the percentage that exceeds 67 kg.
b) To find the weight where 20% are below that weight, we find the Z value for 20% from a Z-table, which is approximately -0.84. Then, use the formula to calculate the weight:
X = Zσ + μ
X = -0.84(9) + 65 = 57.44 kg
Learn more about Z-Score here:
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