The value of seaweed height that divides the bottom 30% from the top 70% is 8.96 cm
<h3>What is standard deviation?</h3>
An indicator of how much a group of numbers vary or are dispersed is the standard deviation. When the standard deviation is low, the values tend to fall within a narrow range of the set's mean, but when it is large, the values tend to deviate from that mean more.
<h3>What is sampling distribution?</h3>
Mean. The population from which the scores were sampled is represented by the mean of the sampling distribution of the mean. As a result, the sampling distribution's mean is also the population's mean if the population as a whole has a mean of.
<h3>According to the given information:</h3>
The height of the seaweed be : x
X~N(μ;σ²)
μ= 10 cm
σ= 2 cm
the value of the variable X at which the distribution's bottom 0.30 and top 0.70 are separated.
P(X≤x)= 0.30
P(X≥x)= 0.70
The Z value that divides the bottom 0.30 from the top 0.70 of the standard normal distribution must be identified. Using the formula
Z=(X-μ)/σ, the Z value must then be converted to the matching X value.
P(Z≤z)= 0.30
Search for the probability of 0.30 in the table's body to get to the margins where the Z value is formed. Since the distribution's mean is "0," negative values can be found in the lower half of the distribution.
z= -0.52
The value of X must now be cleared:
Z= (X-μ)/σ
Z*σ= X-μ
X= (Z*σ)+μ
X= (-0.52*2)+10= 8.96
The value of seaweed height that divides the bottom 30% from the top 70% is 8.96 cm.
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I understand that the question you are looking for is:
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.
