5 times the sum of a number and 4 equals 9

Thickness measurement ancient prehistoric Native American Pot Shards discovered in Hopi Village are approximately normally distr

beks73 [17]

Answer:

a) $P(X$

And we can find this probability using the normal standard table and we got:

$P(z$

b) $P(X>7)=P(\frac{X-\mu}{\sigma}>\frac{7-\mu}{\sigma})=P(Z>\frac{7-5.1}{0.9})=P(z>2.11)$

And we can find this probability using the complement rule and the normal standard table and we got:

$P(z>2.11)=1-P(Z$

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(5.1,0.9)$

Where $\mu=5.1$ and $\sigma=0.9$

Part a

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability using the normal standard table and we got:

$P(z$

Part b

We are interested on this probability

$P(X>7)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>7)=P(\frac{X-\mu}{\sigma}>\frac{7-\mu}{\sigma})=P(Z>\frac{7-5.1}{0.9})=P(z>2.11)$

And we can find this probability using the complement rule and the normal standard table and we got:

$P(z>2.11)=1-P(Z$

4 0

3 years ago

Suppose a sample of 972972 tenth graders is drawn. Of the students sampled, 700700 read above the eighth grade level. Using the

max2010maxim [7]

Answer:

The 85% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.259, 0.301).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

972 students, 700 read above the eight grade level. We want the confidence interver for the proportion of those who read at or below the 8th grade level. 972 - 700 = 272, so $n = 972, \pi = \frac{272}{972} = 0.28$