Find ECD E -D C 31° A B

Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 117 mg/dl and

TiliK225 [7]

Answer:

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

Step-by-step explanation:

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

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 117, \sigma = 10.6, n = 6, s = \frac{10.6}{\sqrt{6}} = 4.33$

What is the level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L ?

This is the value of X when Z has a pvalue of 1-0.01 = 0.99. So X when Z = 2.33.

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

By the Central Limit Theorem

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

$2.33 = \frac{X - 117}{4.33}$

$X - 117 = 2.33*4.33$

$X = 127.1$

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

6 0

3 years ago

Mr. and Mrs. Storey drove 3,200 miles in all during their vacation. Mr. Storey drove 3 times as much as Mrs. Storey. How many mi

baherus [9]

Mr. drove 3 times more then Mrs.. ok

so how far Mrs. drove will be X
and Mr. will be Y
but we know that Mr. drove 3 times more then Mrs therefor
Y= 3X

and the Sum = 3,200

3,200=X+Y
3,200= X+(3X)
3,200=4X
div both by 4
800=X
now we know how far Mrs drove now to find Mr.
again "Mr. drove 3 times more then Mrs"
800 * 3 = 2,400

4 0

3 years ago

To prepare for a bake-off, Clarissa used 560 grams of flour each day for 3 weeks. How many kilograms did Clarissa use in those 3

Sindrei [870]

Convert G to kg
560/1000=.56 kg a day

If the bake off was 5 days
.56*5= 2.8kg... per week
2.8kg*3= 8.4 kg for 3 weeks

If it was 7 days
.56*7 = 3.92 kg per week
3.92*3= 11.76 kg for 3 weeks

5 0

3 years ago

In an examination of purchasing patterns of shoppers, a sample of 16 shoppers revealed that they spent, on average, $54 per hour

Lesechka [4]

Answer:

$54-1.64\frac{21}{\sqrt{16}}=45.39$

$54 +1.64\frac{21}{\sqrt{16}}=62.61$

So on this case the 90% confidence interval would be given by (45.39;62.61)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=54$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=21$ represent the sample standard deviation

n=16 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.90 or 90%, the value of $\alpha=1-0.9=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=1.64$

Now we have everything in order to replace into formula (1):

$54-1.64\frac{21}{\sqrt{16}}=45.39$

$54 +1.64\frac{21}{\sqrt{16}}=62.61$

So on this case the 90% confidence interval would be given by (45.39;62.61)

4 0

3 years ago

The sequence an = 8, 13, 18, 23, ... is the same as the sequence a1 = 8, an = an-1 + 5.

harkovskaia [24]

Actually, this answer would be true. Why?

The first equation is: a(sub n) = 8, 13, 18, 23

The second is: a(sub 1)=8 ; a(sub n)= a(sub n-1)+5
if you wish to find the second term, plug two into the equation for n
8+5=13
to find the third, plug the second term, 13, in for n.
13+5=18.

Hope this helped! I know it's a bit on the late side, but at least you can get the general idea!

8 0

3 years ago

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