the height of seaweed of all plants in a body of water are normally distributed with a mean of 10 cm and a standard deviation of 2 cm. which length separates the lowest 30% of the means of the plant heights in a sampling distribution of sample size 15 from the highest 70%? round your answer to the nearest hundredth. use the z-table below:

Data in a normal distribution are symmetrically distributed and have no skew. The majority of values cluster around a central region, with values decreasing as one moves away from the center. In a normal distribution, the measures of central tendency (mean, mode, and median) are all the same.

Given data:

X: height of seaweed.

X~N (μ;σ²)

μ= 10 cm

σ= 2 cm

We have to find the value of the variable X that separates the bottom 0.30 of the distribution from the top 0.70

P(X ≤ x) = 0.30

P(X ≥ x) = 0.70

Now by using the standard normal distribution,

we have to find the value of Z that separates the bottom 0.30 from the top 0.70 and then use the formula

Z = (X - μ)/σ

translates the Z value to the corresponding X value.

P(Z ≤ z) = 0.30

In the body of the table look for the probability of 0.30 and reach the margins to form the Z value. The mean of the distribution is "0" so below 50% of the distribution you'll find negative values.

z= -0.52

Now you have to clear the value of X:

Z= (X - μ)/σ

X= (Z * σ) + μ

X = (-0.52 * 2) + 10

= 8.96

hence, the value of seaweed height that divides the bottom 30% from the top 70% is 8.96 cm.

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