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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:
0.00 0.01 0.02 0.030.04 0.05 0.06 0.08 0.09 0.07 -0.8 0.212 0.209 0.206 0.203 0.201 0.198 0.195 0.192 0.189 0.187 -0.7 0.242 0.239 0.236 0.233 0.230 0.227 0.224 0.221 0.218 0.215 -0.6 0.274 0.271 0.268 0.264 0.261 0.258 0.255 0.251 0.248 0.245 -0.5 0.309 0.305 0.302 0.298 0.295 0.291 0.288 0.284 0.281 0.278 -0.4 0.345 0.341 0.337 0.334 0.330 0.326 0.323 0.319 0.316 0.312 -0.3 0.382 0.378 0.374 0.3710.367 0.363 0.359 0.356 0.352 0.348
Round the z-score and i to two decimal places. Provide your answer below: Z-Score =

Respuesta :

cchilabert

Answer:

Step-by-step explanation:

Hello!

The variable of interest is:

X: height of seaweed.

X~N(μ;σ²)

μ= 10 cm

σ= 2 cm

You 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

Using the standard normal distribution you have to find the value of Z that separates the bottom 0.30 from the top 0.70 and then using the formula Z= (X-μ)/σ translate 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

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

I hope this helps!

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