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Troyanec [42]
1 year ago
7

The mean life of a television set is 119 months with a standard deviation of 13 months. If a sample of 67 televisions is randoml

y selected, what is the probability that the sample mean would differ from the true mean by less than 2 months? Round your answer to four decimal places.

Mathematics
1 answer:
PIT_PIT [208]1 year ago
5 0

Answer:

Provided in the picture below.

Step-by-step explanation:

Provided in the picture below.

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Rearrange x=3g+2 to make g the subject
Eduardwww [97]

Answer:

g = (x - 2)/3

Step-by-step explanation:

g = (x - 2)/3

8 0
2 years ago
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A circle with radius of \greenD{5\,\text{cm}}5cmstart color #1fab54, 5, start text, c, m, end text, end color #1fab54 sits insid
Marat540 [252]

Answer:

The area of the shaded region is 42.50 cm².

Step-by-step explanation:

Consider the figure below.

The radius of the circle is, <em>r</em> = 5 cm.

The sides of the rectangle are:

<em>l</em> = 11 cm

<em>b</em> = 11 cm.

Compute the area of the shaded region as follows:

Area of the shaded region = Area of rectangle - Area of circle

                                            =[l\times b]-[\pi r^{2}]\\\\=[11\times 11]-[3.14\times 5\times 5]\\\\=121-78.50\\\\=42.50

Thus, the area of the shaded region is 42.50 cm².

7 0
2 years ago
Read 2 more answers
Two different samples will be taken from the same population of test scores where the population mean and standard deviation are
Alenkinab [10]

Answer:

The sample consisting of 64 data values would give a greater precision.

Step-by-step explanation:

The width of a (1 - <em>α</em>)% confidence interval for population mean μ is:

\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{n}}

So, from the formula of the width of the interval it is clear that the width is inversely proportion to the sample size (<em>n</em>).

That is, as the sample size increases the interval width would decrease and as the sample size decreases the interval width would increase.

Here it is provided that two different samples will be taken from the same population of test scores and a 95% confidence interval will be constructed for each sample to estimate the population mean.

The two sample sizes are:

<em>n</em>₁ = 25

<em>n</em>₂ = 64

The 95% confidence interval constructed using the sample of 64 values will have a smaller width than the the one constructed using the sample of 25 values.

Width for n = 25:

\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{25}}=\frac{1}{5}\cdot [2\cdot z_{\alpha/2}\cdot \sigma]        

Width for n = 64:

\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{64}}=\frac{1}{8}\cdot [2\cdot z_{\alpha/2}\cdot \sigma]

Thus, the sample consisting of 64 data values would give a greater precision

5 0
1 year ago
Read 2 more answers
A newly hired basketball coach promised a high-paced attack that will put more points on the board than the team’s previously te
puteri [66]

Answer:

a. z = 2.00

Step-by-step explanation:

Hello!

The study variable is "Points per game of a high school team"

The hypothesis is that the average score per game is greater than before, so the parameter to test is the population mean (μ)

The hypothesis is:

H₀: μ ≤ 99

H₁: μ > 99

α: 0.01

There is no information about the variable distribution, I'll apply the Central Limit Theorem and approximate the sample mean (X[bar]) to normal since whether you use a Z or t-test, you need your variable to be at least approximately normal. Considering the sample size (n=36) I'd rather use a Z-test than a t-test.

The statistic value under the null hypothesis is:

Z= X[bar] - μ  = 101 - 99 = 2

σ/√n 6/√36

I don't have σ, but since this is an approximation I can use the value of S instead.

I hope it helps!

7 0
2 years ago
Suppose two different methods are available for eye surgery. The probability that the eye has not recovered in a month is 0.002
umka21 [38]

Answer:

0.4007

Step-by-step explanation:

Let's define the following events:

A: method A is used

B: method B is used

NR: the eye has not recovered in a month

R: the eye is recovered in a month

The probability that the eye has not recovered in a month is 0.002 if method A is used, i.e., P(NR|A) = 0.002, so P(R|A) = 0.998.

When method B is used, the probability that the eye has not recovered in a month is 0.005, i.e., P(NR|B) = 0.005, so P(R|B) = 0.995.

40% of eye surgeries are done with method A, i.e., P(A) = 0.4

60% of eye surgeries are done with method B, i.e., P(B) = 0.6

If an eye is recovered in a month after surgery is done in the hospital, what is the probability that method A was performed? We are looking for P(A|R), then, by Bayes' Formula

P(A|R) = P(R|A)P(A)/(P(R|A)P(A) + P(R|B)P(B)) = 0.998*0.4/(0.998*0.4 + 0.995*0.6) = 0.4007

4 0
2 years ago
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