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2 years ago

The human resources manager at a company records the length in hours, of one shift at work, X. He creates the probabity

Mathematics

1 answer:

faust18 [17]2 years ago

8 0

Answer:

.84

Step-by-step explanation:

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A random sample from a normal population is obtained, and the data are given below. Find a 90% confidence interval for . 114 157

melamori03 [73]

Answer:

$103.160 \leq \sigma \leq 187.476$

And the upper bound rounded to the nearest integer would be 187.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

Data given: 114 157 203 257 284 299 305 344 378 410 421 450 478 480 512 533 545

The confidence interval for the population variance $\sigma^2$ is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case we need to find the sample standard deviation with the following formula:

$s=sqrt{\frac{\sum_{i=1}^{17} (x_i -\bar x)^2}{n-1}}$

And in order to find the sample mean we just need to use this formula:

$\bar x =\frac{\sum_{i=1}^n x_i}{n}$

The sample mean obtained on this case is $\bar x= 362.941$ and the deviation s=132.250

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=17-1=16$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a tabel to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,16)" "=CHISQ.INV(0.95,16)". so for this case the critical values are:

$\chi^2_{\alpha/2}=26.296$

$\chi^2_{1- \alpha/2}=7.962$

And replacing into the formula for the interval we got:

$\frac{(16)(132.250)^2}{26.296} \leq \sigma \leq \frac{(16)(132.250)^2}{7.962}$

$10641.959 \leq \sigma^2 \leq 35147.074$

Now we just take square root on both sides of the interval and we got:

$103.160 \leq \sigma \leq 187.476$

And the upper bound rounded to the nearest integer would be 187.

4 0

2 years ago

21. Given directed line segment WV , find the coordinates of R

Jobisdone [24]

Answer:

$R\left(\frac {4x_1+3x_2}{7} , \frac {4y_1+3y_2}{7}\right)$

Step-by-step explanation:

Let the coordinate of the points W, V and R are $(x_1,y_1), (x_2,y_2),$ and $(x_s,y_s)$ respectively.

The coordinate of the section point, $(x_s,y_s),$ which divides the line joining the two points $(x_1,y_1)$ and $(x_2,y_2)$ in the ration $m:n$ is

$x_s=\frac {nx_1+mx_2}{m+n}$ and

$y_s=\frac {ny_1+my_2}{m+n}$ .

The given ration is, $m:n=3:4$

$R(x_s,y_s)=R\left(\frac {nx_1+mx_2}{m+n} , \frac {ny_1+my_2}{m+n}\right)$

$=R\left(\frac {4x_1+3x_2}{7} , \frac {4y_1+3y_2}{7}\right)$ .

The exact point can be determined by putting the value of the exact coordinate in the above-obtained formula.

6 0

2 years ago

Rona mixes 2 pounds of meat with some chopped vegetables to make a mixture. She divides the mixture into 4 equal portions. Each

AVprozaik [17]

Answer:

First choice: (1/4)(2 + v) = 3; v = 10 pounds of chopped vegetables

Step-by-step explanation:

"2 pounds of meat with some chopped vegetables"

2 + v

"She divides the mixture into 4 equal portions."

(1/4)(2 + v)

"Each portion weighs 3 pounds."

(1/4)(2 + v) = 3

2 + v = 12

v = 10

Answer: (1/4)(2 + v) = 3; v = 10 pounds of chopped vegetables

7 0

2 years ago

The amount of time it takes for a student to complete a statistics quiz is uniformly distributed (or, given by a random variable

topjm [15]

Answer:

(A) 0.15625

(B) 0.1875

(C) Can't be computed

Step-by-step explanation:

We are given that the amount of time it takes for a student to complete a statistics quiz is uniformly distributed between 32 and 64 minutes.

Let X = Amount of time taken by student to complete a statistics quiz

So, X ~ U(32 , 64)

The PDF of uniform distribution is given by;

f(X) = $\frac{1}{b-a}$ , a < X < b where a = 32 and b = 64

The CDF of Uniform distribution is P(X <= x) = $\frac{x-a}{b-a}$

(A) Probability that student requires more than 59 minutes to complete the quiz = P(X > 59)

P(X > 59) = 1 - P(X <= 59) = 1 - $\frac{x-a}{b-a}$ = 1 - $\frac{59-32}{64-32}$ = $1-\frac{27}{32}$ = 0.15625

(B) Probability that student completes the quiz in a time between 37 and 43 minutes = P(37 <= X <= 43) = P(X <= 43) - P(X < 37)

P(X <= 43) = $\frac{43-32}{64-32}$ = $\frac{11}{32}$ = 0.34375

P(X < 37) = $\frac{37-32}{64-32}$ = $\frac{5}{32}$ = 0.15625

P(37 <= X <= 43) = 0.34375 - 0.15625 = 0.1875

= P(X = 44.74)

The above probability can't be computed because this is a continuous distribution and it can't give point wise probability.

3 0

2 years ago

Jeanne wants to start collecting coins and orders a coin collection starter kit. The kit comes with three coins chosen at random

lesya692 [45]

Conditional probability is a measure of the probability of an event given that another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes $P_B(A)$ .

The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by
$P(A|B)= \frac{P(A \cap B)}{P(B)}$

In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.

The probability of selecting one coin is $\frac{1}{3}$

Part A:
To find the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.

P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.

Thus $P(A)= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}$

P(B) means that the first envelope contains a quarter AND the second envelope contains a quarter

Thus $P(B)= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$

Therefore, $P(A|B)=\left( \frac{ \frac{1}{27} }{ \frac{1}{9} } \right)= \frac{1}{3}$

Part B:
To find the probability that all three coins are different if the first envelope Jeanne opens contains a dime, let the event that all three coins are different be C and the event that the first envelope Jeanne opens contains a dime be D.

$P(C)= \frac{3}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{6}{27} = \frac{2}{9}$

 $P(D)= \frac{1}{3}$ 

Therefore, $P(C|D)=\left( \frac{ \frac{2}{9} }{ \frac{1}{3} } \right)= \frac{2}{3}$

3 0

2 years ago