In a distribution exactly normal with mean 47.5 Kilogram, 89.97% of the items are under 70 kilogram weight. Then which of the following is the standard deviation of the distribution? Given that area under standard normal curve from Z=0 to Z=1.28 is 0.3997.*16.578 Kilogram17.578 Kilogram16.875 Kilogram17.875 Kilogram
Question
In a distribution exactly normal with mean 47.5 Kilogram, 89.97% of the items are under 70 kilogram weight. Then which of the following is the standard deviation of the distribution? Given that area under standard normal curve from Z=0 to Z=1.28 is 0.3997.*16.578 Kilogram17.578 Kilogram16.875 Kilogram17.875 Kilogram
Solution
To solve this problem, we need to use the properties of a normal distribution.
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First, we know that 89.97% of the items are under 70 kg. This means that the Z-score for 70 kg is the value at which 89.97% of the distribution lies to the left.
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Given that the area under the standard normal curve from Z=0 to Z=1.28 is 0.3997, and we know that 50% of the distribution lies to the left of the mean in a standard normal distribution, we can add these two values together to find the total percentage of the distribution that lies to the left of Z=1.28. This gives us 0.3997 + 0.5 = 0.8997, or 89.97%.
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Therefore, we know that the Z-score for 70 kg is 1.28.
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The Z-score is calculated using the formula Z = (X - μ) / σ, where X is the value from the distribution, μ is the mean, and σ is the standard deviation. We can rearrange this formula to solve for σ: σ = (X - μ) / Z.
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Substituting the known values into this formula gives us σ = (70 - 47.5) / 1.28 = 17.578 Kilogram.
So, the standard deviation of the distribution is 17.578 Kilogram.
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