We can estimate the expected value of the various percentiles assuming the normal distribution.
The median, 50th percentile, is the mean, 0.785 mm.
Typically we remember ±1 standard deviation is 68% of the probability, so one standard deviation below the mean is 16th percentile (50-68/2) and one standard deviation above the mean is 84th percentile.
The quartiles, 25th and 75th percentile, are about 2/3 of a standard deviation away from the mean. In other words ±.67 standard deviations is 50% of the probability. It's not a commonly remembered number like 68-95-99.7, but perhaps it should be.
The 25th percentile is 0.785 - (.67)(.07) = <span>0.7381 mm.
</span>The 75th percentile is 0.785 + (.67)(.07) = 0.8319 mm.
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-50-15n =g(n-)+
-50. -50
-15n=g(n-)50
-15. -15
N=g(-15n)50
Answer:
X₀ = 71.65
Step-by-step explanation:
given,
mean (μ ) = 64
standard deviation (σ) = 9
σ² = 81
selected candidate are top 20 % candidate
P( X > X₀) = 0.2
P( Z > Z₀) = 0.2
X₀ = 71.65
Hence, the cut off score is equal to X₀ = 71.65
Answer:1:00 am hope its right
Step-by-step explanation:
Answer:
4x^5 – x^4
Step-by-step explanation:
i took it and was correct
