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aleksley [76]
3 years ago
10

Suppose the number of free throws in a basketball game by one player are normally distributed with a standard deviation 0.97 fre

e throws. A random sample of basketball players from the population produces a sample mean of x¯=4.9 free throws. What value of z should be used to calculate a confidence interval with a 95% confidence level? 20.10 1.282 20.05 1.645 0.025 1.960 20.005 2.576 2.326
Mathematics
1 answer:
babunello [35]3 years ago
3 0

Answer: 1.960

Step-by-step explanation:

The value of z we use to calculate a confidence interval with a (1-\alpha) confidence level is a two-tailed test value i.e. represented by :-

              z_{\alpha/2}

Given : The level of confidence: 1-\alpha=0.95

Then, significance level : \alpha: 1-0.95=0.05

With the help of standard normal distribution table for z , we have

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.960

Hence, the value of z should be used to calculate a confidence interval with a 95% confidence level =1.960

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A group of dental researchers are testing the effects of acidic drinks on dental crowns. They have five containers of crowns lab
mestny [16]

Answer:

\begin{array}{cccccc}{x} & {V} & {W} & {X} & {Y} & {Z} & P(x) & {0.20} & {0.20} & {0.20} & {0.20} & {0.20} \ \end{array}

Step-by-step explanation:

Given

S = \{V,W,X,Y,Z\}

n(S) = 5

Required

The probability model

To do this, we simply calculate the probability of each container.

So, we have:

P(V) = \frac{n(V)}{n(S)} = \frac{1}{5} = 0.20

P(W) = \frac{n(W)}{n(S)} = \frac{1}{5} = 0.20

P(X) = \frac{n(X)}{n(S)} = \frac{1}{5} = 0.20

P(Y) = \frac{n(Y)}{n(S)} = \frac{1}{5} = 0.20

P(Z) = \frac{n(Z)}{n(S)} = \frac{1}{5} = 0.20

So, the probability model is:

\begin{array}{cccccc}{x} & {V} & {W} & {X} & {Y} & {Z} & P(x) & {0.20} & {0.20} & {0.20} & {0.20} & {0.20} \ \end{array}

5 0
2 years ago
I don't understand at all. <br><br> Please give me all the details about this thing
DerKrebs [107]
I know that B is orthographic. 
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7 0
3 years ago
Read 2 more answers
-----.135as a fraction
Mrac [35]
This has to do with place value.
0.135.....the last digit (5) is in the thousandths place....so put it over 1000.
135/1000 reduces to 27/200
6 0
3 years ago
A number is written with the following factorization: 22 × 3 × 54 × 8 × 11 2 . Is this factorization a prime factorization? Expl
Flauer [41]

Answer:

This factorization is not a prime factorization.

Step-by-step explanation:

Prime factorization always singles down to prime numbers. The following factorization is not a prime factorization because the factors are not all prime. For example, 22 is not prime.

4 0
3 years ago
A tire manufacturer warranties its tires to last at least 20,000 miles orâ "you get a new set ofâ tires." In itsâ experience, a
Y_Kistochka [10]

Answer:

Probability that a set of tires wears out before 20,000 miles is 0.1151.

Step-by-step explanation:

We are given that a tire manufacturer warranties its tires to last at least 20,000 miles or "you get a new set of tires." In its past experience, a set of these tires last on average 26,000 miles with S.D. 5,000 miles. Assume that the wear is normally distributed.

<em>Let X = wearing of tires</em>

So, X ~ N(\mu=26,000,\sigma^{2}=5,000^{2})

Now, the z score probability distribution is given by;

         Z = \frac{X-\mu}{\sigma} ~ N(0,1)

where, \mu = average lasting of tires = 26,000 miles

            \sigma = standard deviation = 5,000 miles

So, probability that a set of tires wears out before 20,000 miles is given by = P(X < 20,000 miles)

    P(X < 20,000) = P( \frac{X-\mu}{\sigma} < \frac{20,000-26,000}{5,000} ) = P(Z < -1.2) = 1 - P(Z \leq 1.2)

                                                                    = 1 - 0.88493 = 0.1151

Therefore, probability that a set of tires wears out before 20,000 miles is 0.1151.

4 0
3 years ago
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