Question

To evaluate whether or not the intake of a vitamin or mineral is adequate, comparisons are made between the intake distribution and the requirement distribution. Here is some information about the distribution of vitamin C intake, in milligrams per day, for women aged 19 to 30 years:
Percentile (mg/d)
Mean 1st 5th 19th 25th 50th 75th 90th 95th 99th
84.2 31 43 47 60 79 103 126 141 180

Use the 5th, the 50th, and the 95th percentiles of this distribution to estimate the mean (±0.01) and standard deviation (±0.01) of this distribution assuming that the distribution is Normal.

μ = _________
σ = _________

Answer 1

Answer:

$Mean=79$

$\sigma=30.3951.$

Step-by-step explanation:

From the question we are told that:

$Age Bracket :19-20$

$5th\ percentile = 42$  

$50th\ percentile = 79$  

$95th\ percentile = 142.$

Generally the mean Median and mode of the 50th percentile is are all equal

$Mean=Median=Mode$

Therefore

$Mean=79$

Generally for Normal distribution

$5th\ percentile\ = mean - 1.645*\sigma$  

$95th\ percentile\ = mean + 1.645*\sigma$

Therefore

$(95th\ percentile\ - 5th\ percentile) = 2*(1.645*SD).$

$\sigma=(95th\ percentile\ - 5th\ percentile)/3.29$

$\sigma=\frac{142-42}{3.29}$

$\sigma=30.3951.$