Solve the proportion<br> h/4=7/14

I need help I’m having trouble

Svet_ta [14]

They would be 8cm far apart.

3 0

3 years ago

Problem: The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72

Lisa [10]

Answer:

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

1) 0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2) 0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3) 0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4) 0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

Step-by-step explanation:

To solve these questions, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72 inches and standard deviation 3.17 inches.

This means that $\mu = 38.72, \sigma = 3.17$

Sample of 10:

This means that $n = 10, s = \frac{3.17}{\sqrt{10}}$

Compute the probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

This is 1 subtracted by the p-value of Z when X = 40. So

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

By the Central Limit Theorem

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

$Z = \frac{40 - 38.72}{\frac{3.17}{\sqrt{10}}}$

$Z = 1.28$

$Z = 1.28$ has a p-value of 0.8997

1 - 0.8997 = 0.1003

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

$\mu = 266, \sigma = 16$

1. What is the probability a randomly selected pregnancy lasts less than 260 days?

This is the p-value of Z when X = 260. So

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

$Z = \frac{260 - 266}{16}$

$Z = -0.375$

$Z = -0.375$ has a p-value of 0.3539.

0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less?

Now $n = 20$ , so:

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

$Z = \frac{260 - 266}{\frac{16}{\sqrt{20}}}$

$Z = -1.68$

$Z = -1.68$ has a p-value of 0.0465.

0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less?

Now $n = 50$ , so:

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

$Z = \frac{260 - 266}{\frac{16}{\sqrt{50}}}$

$Z = -2.65$

$Z = -2.65$ has a p-value of 0.0040.

0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?

Sample of size 15 means that $n = 15$ . This probability is the p-value of Z when X = 276 subtracted by the p-value of Z when X = 256.

X = 276

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

$Z = \frac{276 - 266}{\frac{16}{\sqrt{15}}}$

$Z = 2.42$

$Z = 2.42$ has a p-value of 0.9922.

X = 256

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

$Z = \frac{256 - 266}{\frac{16}{\sqrt{15}}}$

$Z = -2.42$

$Z = -2.42$ has a p-value of 0.0078.

0.9922 - 0.0078 = 0.9844

0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

8 0

3 years ago

If an intravenous solution containing 123 mg of a drug substance in each 250-mL bottle is to be administered at the rate of 200

12345 [234]

Answer:

24.39mL of the solution would be given per hour.

Step-by-step explanation:

This problem can be solved by direct rule of three, in which there are a direct relationship between the measures, which means that the rule of three is a cross multiplication.

The first step to solve this problem is to see how many mg of the solution is administered per hour.

Each minute, 200 ug are administered. 1mg has 1000ug, so

1mg - 1000 ug

xmg - 200 ug

$1000x = 200$