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3 years ago

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Assume that the weight of two year old babies have distribution that is approximately normal with a mean of 29 pounds and a stan

dard deviation of 3 pounds. what weight of two year old baby corresponds to 10th percentile?

1 answer:

Lana71 [14]3 years ago

5 0

Answer:

25.15 ponds is the weight of two year old baby corresponds to 10th percentile.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 29 pounds

Standard Deviation, σ = 3 pounds

We are given that the distribution of weight of two year old babies is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.10

P(X < x)

$P( X < x) = P( z < \displaystyle\frac{x - 29}{3})=0.10$

Calculation the value from standard normal z table, we have,

$P(z < -1.282) = 0.10$

$\displaystyle\dfrac{x - 29}{3} = -1.282\\x = 25.154 \approx 25.15$

Thus, 25.15 ponds is the weight of two year old baby corresponds to 10th percentile.

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Step-by-step explanation:

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<u>Substitute the value of </u> $x$ <u> into the 2nd equation and solve for </u> $z$ <u>:</u>

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<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

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The proportion of sample means greater than 8.8 hours is <u>one subtracted by the p-value of Z when X = 8.8</u>, hence:

$Z = \frac{X - \mu}{\sigma}$

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More can be learned about the normal distribution at brainly.com/question/25800303

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