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vitfil [10]
2 years ago
7

A forestry researcher wants to estimate the average height of trees in a forest near Atlanta, Georgia. She takes a random sample

of 18 trees from this forest. The researcher found that the average height was 4.8 meters with a standard deviation of 0.55 meters. Assume that the distribution of the heights of these trees is normal. For this sample what is the margin of error for her 99% confidence interval
Mathematics
1 answer:
marusya05 [52]2 years ago
7 0

Answer:

The margin of error for her confdence interval is of 0.3757.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 18 - 1 = 17

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 17 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.99}{2} = 0.995. So we have T = 2.8982

The margin of error is:

M = T\frac{s}{\sqrt{n}}

In which s is the standard deviation of the sample and n is the size of the sample.

Standard deviation of 0.55 meters.

This means that s = 0.55

What is the margin of error for her 99% confidence interval?

M = T\frac{s}{\sqrt{n}}

M = 2.8982\frac{0.55}{\sqrt{18}}

M = 0.3757

The margin of error for her confdence interval is of 0.3757.

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Suppose that you bet $5 on each of a sequence of 50 independent fair games. Use the central limit theorem to approximate the pro
amm1812

Answer:

chances chances of happening = 0.0119

Step-by-step explanation:

given data

bet =  $5

independent fair games = 50

solution

we will think game as the normal distribution

so here  mean is will be

mean = \frac{50}{2}

mean = 25

and standard deviation will be

standard deviation = \sqrt{50*0.5*0.5}

standard deviation = 3.536

so

we have to lose 33 out of 50 time for lose more than $75

so as chance of doing things z score is

z score = \frac{33-25}{3.536}

z score  = 2.26

so from z table

chances chances of this happening = 0.0119

8 0
2 years ago
If c (x) = StartFraction 5 Over x minus 2 EndFraction and d(x) = x + 3, what is the domain of (cd)(x)?
Volgvan

Answer:

The domain of the function (cd)(x) will be all real values of x except x = 2.

Step-by-step explanation:

The two functions are c(x) = \frac{5}{x - 2} and d(x) = x + 3

So, (cd)(x) = (\frac{5}{x - 2})(x + 3) = \frac{5(x + 3)}{x - 2}

Then, for x = 2 the function (cd)(x) will be undefined as zero in the denominator will make the function (cd)(x) undefined.

Therefore, the domain of the function (cd)(x) will be all real values of x except x = 2. (Answer)

8 0
2 years ago
Read 2 more answers
Two students from a group of eight boys and 12 girls are sent to represent the school in a parade. If the students are chosen at
kari74 [83]

Answer:

Step-by-step explanation:

* Lets explain how to find the probability of an event  

- The probability of an Event = Number of favorable outcomes ÷ Total

 number of possible outcomes

- P(A) = n(E) ÷ n(S) , where

# P(A) means finding the probability of an event A  

# n(E) means the number of favorable outcomes of an event

# n(S) means set of all possible outcomes of an event

- Probability of event not happened = 1 - P(A)

- P(A and B) = P(A) . P(B)

* Lets solve the problem

- There is a group of students

- There are 8 boys and 12 girls in the group

∴ There are 8 + 12 = 20 students in the group

- The students are sent to represent the school in a parade

- Two students are chosen at random

∴ P(S) = 20

- The students that chosen are not both girls

∴ The probability of not girls = 1 - P(girls)

∵ The were 20 students in the group

∵ The number of girls in the group was 12

∴ The probability of chosen a first girl = 12/20

∵ One girl was chosen, then the number of girls for the second

  choice is less by 1 and the total also less by 1

∴ The were 19 students in the group

∵ The number of girls in the group was 11

∴ The probability of chosen a second girl = 11/19

- The probability of both girls is P(1st girle) . P(2nd girl)

∴ The probability of both girls = (12/20) × (11/19) = 33/95

- To find the probability of both not girls is 1 - P(both girls)

∴ P(not both girls) = 1 - (33/95) = 62/95

* The probability that the students chosen are not both girls is 62/95

3 0
2 years ago
Read 2 more answers
The mean hourly wage for employees in goods-producing industries is currently $24.57 (Bureau of Labor Statistics website, April,
sweet-ann [11.9K]

Answer:

a. H_{0}: μ = $24.57

H_{a}: μ ≠ $24.57

b. 0.12

c. we fail to reject the null hypothesis since p ≈0.12 > 0.05

d. we fail to reject the null hypothesis since z ≈-1.6 > -1.96

Step-by-step explanation:

a)

H_{0}: μ = $24.57

H_{a}: μ ≠ $24.57

b)

To compute p value, we need to calculate z-score of $23.89 per hour. in the distribution assumed under null hypothesis

z-score can be calculated as follows:

\frac{X-M}{\frac{s}{\sqrt{N} } } where

  • X= $23.89
  • M is the mean value under null hypothesis ( $24.57)
  • s is the sample standard deviation ($2.40)
  • N is the sample size (30)

Putting the numbers in the formula we get

\frac{23.89-24.57}{\frac{2.40}{\sqrt{30} } } ≈ −1,5518

The corresponding p-value is <em>two tailed </em>and  ≈ 0.12

c)

since p ≈0.12 > 0.05 <em>we fail to reject the null hypothesis.</em>

d)

Critical value for the 0.05 significance level in two tailed test is :

c(z)=-1.96 Since z≈ −1,6 >-1.96, it is not in the critcal region, therefore <em>we fail to reject the null hypothesis. </em>

7 0
2 years ago
Let $\overline{XY}$ be a tangent to a circle, and let $\overline{XBA}$ be a secant of the circle, as shown below. If $AX = 15$ a
KiRa [710]

Answer:

AB=5.4 units

Step-by-step explanation:

We using the theorem of intersecting tangent and secant to solve this.

By this theorem:

XY^2=XB \times AX\\9^2=15(15-XB)\\81=225-15XB\\15XB=144\\XB=9.6\\$Therefore:\\AB=AX-XB\\AB=15-9.6\\AB=5.4$ units

8 0
2 years ago
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