Please answer quickly! will give brainliest!

The time taken to prepare the envelopes to mail a weekly report to all executives in a company has a normal distribution, with a

telo118 [61]

Answer:

Mailing preparation takes 38.29 min max time to prepare the mails.

Step-by-step explanation:

Given:

Mean:35 min

standard deviation:2 min

and 95% confidence interval.

To Find:

In normal distribution mailing preparation time taken less than.

i.eP(t<x)=?

Solution:

Here t -time and x -required time

mean time 35 min

5 % will not have true mean value . with 95 % confidence.

Question is asked as ,preparation takes less than time means what is max time that preparation will take to prepare mails.

No mail take more time than that time .

by Z-score or by confidence interval is

Z=(X-mean)/standard deviation.

Z=1.96 at 95 % confidence interval.

1.96=(X-35)/2

3.92=(x-35)

X=38.29 min

or

Confidence interval =35±Z*standard deviation

=35±1.96*2

=35±3.92

=38.29 or 31.71 min

But we require the max time i.e 38.29 min

And by observation we can also conclude the max time from options as 38.29 min.

4 0

4 years ago

What is the answer to 1-r=-5

ANTONII [103]

The answer is 6 = r.

Work:
1 - r + r = -5 + r
1 + 5 = -5 + r + 5
6 = r

3 0

4 years ago

It is desired to determine whether there is less variability in the silver plating done by Company 1 than in that done by Compan

uysha [10]

Answer:

The conclusion is True

Step-by-step explanation:

Solution:

- The standard deviation of company s1 = 0.035

- The standard deviation of company s2 = 0.062

- Null Hypothesis : s1^2 = s2^2

- Alternate hypothesis : s1^2 < s2^2

- Criteria to reject Null: M > M_a ( 12 - 1 , 12 - 1 )

- From Tables, M > 2.82

- M - statistics value M = s2^2 / s1^2

- M = (0.062 / 0.035)^2 = 3.14

- M = 3.14 > 2.82 ... Criteria of rejection is met

- Null Hypothesis must be rejected at level of significance a = 0.05

- p value, 1 - 0.965 = 0.035 < 0.05 ( Significance Level )

- The conclusion is True

8 0

4 years ago

Suppose an article reported that 16% of unmarried couples in the United States are mixed racially or ethnically. Consider the po

makkiz [27]

Answer:

39.36% probability that the proportion of couples in the sample who are racially or ethnically mixed will be greater than 0.17.

Step-by-step explanation:

For each couple, there are only two possible outcomes. Either they are mixed racially or ethnically, or they are not. This means that the binomial probability distribution is used to solve this problem.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$n = 100, p = 0.16$

So

$E(X) = 100*0.16 = 16$

$\sqrt{Var(X)} = \sqrt{100*0.16*0.84} = 3.67$

When n = 100, what is the probability that the proportion of couples in the sample who are racially or ethnically mixed will be greater than 0.17?

This is 1 subtracted by the pvalue of Z when $X = 100*0.17 = 17$ . So:

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

$Z = \frac{17 - 16}{3.67}$

$Z = 0.27$

$Z = 0.27$ has a pvalue of 0.6064.

So there is a 1-0.6064 = 0.3936 = 39.36% probability that the proportion of couples in the sample who are racially or ethnically mixed will be greater than 0.17.

4 0

3 years ago

Mrs. Nestler enjoys listening to classical music. She has the following audio CDs by her favorite composers in her collection: 4

ivolga24 [154]

Answer:

10.55% probability

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the CDs are chosen is not important. So we use the combinations formula to solve this question.

1 Bach CD, from a set of 4.

1 Beethoven CD, from a set of 6.

1 Brahms CD, from a set of 3.

1 Handel CD, from a set of 2.

So, D=144

4 CDs from a set of 4+6+3+2 = 15.

So, T= 1365

p= D/T= 144/1365 = 0.1055

10.55% probability that she will choose one by each composer

4 0

3 years ago