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Anestetic [448]

3 years ago

8

Of the 64 people who answered yes to question 6 were male. of the 70 people that answered no to the question 8 or male. if one p

erson is selected at random from the group what is the probability that the person answered yes or was male?
0.537
0.582
0.104
0.094

1 answer:

Colt1911 [192]3 years ago

3 0

Answer:

0.537

Step-by-step explanation:

There were 134 total people that were asked the question.

64 people said yes.

8 who said no were male.

So, 72 of the 134 said yes or were male.

The equation is 72/134.

72/134 ≈ 0.537!

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It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So $\mu = 375, \sigma = 68$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So $n = 6, s = \frac{68}{\sqrt{6}} = 27.76$

This probability is 1 subtracted by the pvalue of Z when X = 410.

$Z = \frac{X - \mu}{s}$

$Z = \frac{410 - 375}{27.76}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So $\mu = 522, \sigma = 106$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?

So $n = 6, s = \frac{106}{\sqrt{6}} = 43.27$

This probability is 1 subtracted by the pvalue of Z when X = 410.

$Z = \frac{X - \mu}{s}$

$Z = \frac{410 - 523}{43.27}$

$Z = -2.61$

$Z = -2.61$ has a pvalue of 0.0045.

So there is a 1-0.0045 = 0.9955 = 99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

6 0

3 years ago

Evaluate the expression for the given value of the variable. <br> 29 + 5/6r <br> r = -1/2<br> find r

Dmitrij [34]

Hey there!

The solution to your question is

We simply plug the "r" value into the equation and solve:

29 + (5/6)(-1/2)

29 + -5/12

28 7/12

Hope it helps! Have an amazing day, and remember, You've got this!

4 0

2 years ago

Bob Nale is the owner of Nale's Texaco GasTown. Bob would like to estimate the mean number of litres (L) of gasoline sold to his

Fudgin [204]

Answer:

a

The point estimate of the population mean is $\= x = 56$

b

The 80% confidence level is $50.57 < \mu < 61.43$

c

There is 80% confidence that the true population mean lies within the confidence interval.

Step-by-step explanation:

From the question we are told that

The sample size is n = 18

The standard deviation is $\sigma = 18 \ L$

The sample mean is $\= x = 56$

Generally the point estimate of the population mean is equivalent to the sample mean whose value is $\= x = 56$

Given that the confidence interval is 80% then the level of significance is mathematically represented as

$\alpha = 100 - 80$

$\alpha = 20 \%$

$\alpha = 0.20$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

The value is $Z_{\frac{ \alpha }{2} } = 1.28$

Generally the margin of error is mathematically evaluated as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E = 1.28 * \frac{18 }{\sqrt{18} }$

=> $E = 5.43$

Generally the 80% confidence interval is mathematically represented as

$\= x - E < \mu < \= x + E$

=> $56 - 5.43 < \mu < 56 + 5.43$

=> $50.57 < \mu < 61.43$

The interpretation is that there is 80% confidence that the true population mean lies within the limit

3 0

3 years ago

Simplify: -2 1/3 - (-10 1/6) 1) -12 1/2 2) 12 1/2 3) 7 5/6 4) -7 5/6

marshall27 [118]

Answer:

- 5 16/22

Step-by-step explanation:

idk

7 0

3 years ago

What is the domain of the function g(x)=5^2^x

zubka84 [21]

$g(x)=5^{2^x}\\\\The\ domain:\ all\ real\ numbers\\\\D=\mathbb{R}$

3 0

3 years ago