A man is on exit 64 at 12:13 he arrived at exit 148 at1:33 was he speeding

The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g

Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

Step-by-step explanation:

This is a problem of the distribution of sample means. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves normally for samples sizes equal or greater than 30 $\\ n \geq 30$ . Mathematically

$\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a normal standard distribution, i.e., a normal distribution that has a population mean $\\ \mu = 0$ and a population standard deviation $\\ \sigma = 1$ .

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

From the question, we know that

The population mean is $\\ \mu = 88$ WPM
The population standard deviation is $\\ \sigma = 14$ WPM

We also know the size of the sample for this case: $\\ n = 137$ sixth graders.

We need to estimate the probability that a sample mean being greater than $\\ \overline{X} = 89.87$ WPM in the distribution of sample means. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}$

$\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}$

$\\ Z = 1.5634 \approx 1.56$

This is a standardized value and it tells us that the sample with mean 89.87 is 1.56 standard deviations above the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any cumulative standard normal table available in Statistics books or on the Internet. Of course, this probability is the same that $\\ P(\overline{X} < 89.87)$ . Then

$\\ P(z$

However, we are looking for P(z>1.56), which is the complement probability of the previous probability. Therefore

$\\ P(z>1.56) = 1 - P(z$

$\\ P(z>1.56) = P(\overline{X}>89.87) = 0.0594$

Thus, "The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

5 0

3 years ago

Draw a rhombus that is not a square. Label the lengths of its sides and its height so it represents a rhombus with an area of 20

Tcecarenko [31]

Answer:

/\

$\sqrt{20}$ / \ $\sqrt{20}$

/ \

\ /

$\sqrt{20}$ \ / $\sqrt{20}$

\/

Step-by-step explanation:

i hope this is right it took me 30 min for the ascii

6 0

3 years ago

A bread company makes 9 types of bread. A restaurant chain ordered 497 loaves of each kind of bread. How many loaves of bread in

Aleksandr [31]

9 types of bread....a restaurant orders 497 loaves of each type

so there are (497 * 9) = 4473 loaves ordered in total

7 0

3 years ago

A bakery packages 868 cupcakes into 31 boxes. The same number of cupcakes are pput into each box. How many cupcakes are in each

ratelena [41]

To find this you would divide 868cupcakes by 31boxes witches is 28cupcakes in each box
hope this helps good luck have a good nite hope u enjoy ur the rest of ur weekend

6 0

3 years ago

Find cot theta. (-6,-3sqrt(5)

cupoosta [38]

0 Pretend that is theta XD

cot 0 = $\frac{1}{tan0}$