Answer:
We conclude that the population mean is equal to 490.
Step-by-step explanation:
We are given that a random sample of 15 observations is selected from a normal population. The sample mean was 495 and the sample standard deviation 9.
Let [tex]\mu[/tex] = population mean.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 490 {means that the population mean is equal to 490}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu\neq[/tex] 490 {means that the population mean is different from 490}
The test statistics that will be used here is One-sample t-test statistics because we don't know about population standard deviation;
T.S. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_1_4[/tex]
where, [tex]\bar X[/tex] = sample mean = 495
s = sample standard deviation = 9
n = sample of observations = 15
So, the test statistics = [tex]\frac{495-490}{\frac{9}{\sqrt{15} } }[/tex] ~ [tex]t_1_4[/tex]
= 2.152
The value of t-test statistics is 2.152.
Now, at a 0.01 level of significance, the t table gives a critical value of -2.977 and 2.977 at 14 degrees of freedom for the two-tailed test.
Since the value of our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as the test statistics will not fall in the rejection region.
Therefore, we conclude that the population mean is equal to 490.
The energy E (in ergs) released by an earthquake is approximated by log E= 11.8 + 1.5M. Where M is the magnitude of the earthquake. What is the energy released by the 1906 San Francisco quake, which measured 8.3 on the Richter scale? This energy, it is estimated, would be sufficient to provide the entire world's food requirements for a day. Answer in ergs.
Answer:
[tex]\large \boxed{3.4 \times 10^{10}\text{ ergs }}[/tex]
Step-by-step explanation:
[tex]\begin{array}{rcl}\log E & = & 11.8 + 1.5M\\& = & 11.8 + 1.5 \times 8.3\\& = & 11.8 + 12.45\\& = & 24.25\\E & = & e^{24.25}\\& = & \mathbf{3.4 \times 10^{10}} \textbf{ ergs}\\\end{array}\\\text{ The energy released was $\large \boxed{\mathbf{3.4 \times 10^{10}}\textbf{ ergs }}$}[/tex]
An investigator claims, with 95 percent confidence, that the interval between 10 and 16 miles includes the mean commute distance for all California commuters. To have 95 percent confidence signifies that
Answer:
Hello the options to your question is missing below are the options
A) if sample means were obtained for a long series of samples, approximately 95 percent of all sample means would be between 10 and 16 miles
B.the unknown population mean is definitely between 10 and 16 miles
C.if these intervals were constructed for a long series of samples, approximately 95 percent would include the unknown mean commute distance for all Californians
D.the unknown population mean is between 10 and 16 miles with probability .95
Answer : if these intervals were constructed for a long series of samples, approximately 95 percent would include the unknown mean commute distance for all Californians ( c )
Step-by-step explanation:
95% confidence
interval = 10 to 16 miles
To have 95% confidence signifies that if these intervals were constructed for a long series of samples, approximately 95 percent would include the unknown mean commute distance for all Californians
confidence interval covers a range of samples/values in the interval and the higher the % of the confidence interval the more precise the interval is,
Three-fourths (x minus 8) = 12
Answer:
x=24
Step-by-step explanation:
3/4(x-8)=12
3/4x-24/4=12
3/4x=18
18 dived by 3/4
x=24
your welcome :)
Given v(x) = g(x) (3/2*x^4 + 4x – 1), find v'(2).
Answer:
Step-by-step explanation:
Given that v(x) = g(x)×(3/2*x^4+4x-1)
Let's find V'(2)
V(x) is a product of two functions
● V'(x) = g'(x)×(3/2*x^4+4x-1)+ g(x) ×(3/2*x^4+4x-1)
We are interested in V'(2) so we will replace x by 2 in the expression above.
g'(2) can be deduced from the graph.
● g'(2) is equal to the slope of the tangent line in 2.
● let m be that slope .
● g'(2) = m =>g'(2) = rise /run
● g'(2) = 2/1 =2
We've run 1 square to the right and rised 2 squares up to reach g(2)
g(2) is -1 as shown in the graph.
■■■■■■■■■■■■■■■■■■■■■■■■■■
Let's derivate the second function.
Let h(x) be that function
● h(x) = 3/2*x^4 +4x-1
● h'(x) = 3/2*4*x^3 + 4
● h'(x) = 6x^3 +4
Let's calculate h'(2)
● h'(2) = 6 × 2^3 + 4
● h'(2) = 52
Let's calculate h(2)
●h(2) = 3/2*2^4 + 4×2 -1
●h(2)= 31
■■■■■■■■■■■■■■■■■■■■■■■■■■
Replace now everything with its value to find V'(2)
● V'(2) = g'(2)×h(2) + g(2)× h'(2)
● V'(2)= 2×31 + (-1)×52
●V'(2) = 61 -52
●V'(2)= 9
Sarah has $30,000 in her bank account today. Her grand-father has opened this account for her 15 years ago when she was born. Calculate the money that was deposited in the account 15 years ago if money has earned 3.5% p.a. compounded monthly through all these years.
Answer:
Deposit value(P) = $17,760 (Approx)
Step-by-step explanation:
Given:
Future value (F) = $30,000
Number of Year (n) = 15 year = 15 × 12 = 180 month
rate of interest (r) = 3.5% = 0.035 / 12 = 0.0029167
Find:
Deposit value(P)
Computation:
[tex]A = P(1+r)^n\\\\ 30000 = P(1+0.0029167)^{180} \\\\ 30000 = P(1.68917) \\\\ P = 17760.2018[/tex]
Deposit value(P) = $17,760 (Approx)
Yo help me real quick?
Answer:
1,2 and 6
Step-by-step explanation:
pie symbol
2/3
0.333333....
What is the difference between a line graph and a scatter plot?
Step-by-step explanation:
scatter plot s are similar to line graphs in that they start with mapping quantitive data points. The difference is that with a scatter plot, the decision is made the the individual points should not be connected directly together with a line but, instead express a trend
What is credit?
an arrangement in which you receive money, goods, or services now in exchange for the promise of payment later
an arrangement in which you receive goods or services in exchange for other goods and services
an arrangement in which you receive money now and pay it bulk later with fees?
A news article estimated that only 5% of those age 65 and older who prefer to watch the news, rather than to read or listen, watch the news online. This estimate was based on a survey of a large sample of adult Americans. Consider the population consisting of all adult Americans age 65 and older who prefer to watch the news, and suppose that for this population the actual proportion who prefer to watch online is 0.05. A random sample of n = 100 people will be selected from this population and p, the proportion of people who prefer to watch online, will be calculated.
(a) What are the mean and standard deviation of the sampling distribution of p? (Round your standard deviation to four decimal places.
(b) Is the sampling distribution of p approximately normal for random samples of size n 100? Explain.
i. The sampling distribution of p is approximately normal because np is less than 10.
ii. The sampling distribution of p is approximately normal because np is at least 10.
iii. The sampling distribution of p is not approximately normal because np is less than 10
iv. The sampling distribution of p is not approximately normal because np is at least 10
v. The sampling distribution of p is not approximately normal because n(1 - p) is less than 10.
(c) Suppose that the sample size is n = 400 rather than n = 100, what are the values for the mean and standard deviation when n=400?
Does the change in sample size affect the mean and standard deviation of the sampling distribution of p? If not, explain why not.
i. When the sample size increases, the mean increases.
ii. When the sample size increases, the mean decreases.
iii. When the sample size increases, the mean stays the same.
iv. The sampling distribution is always centered at the population mean, regardless of sample size.
v. When the sample size increases, the standard deviation increases.
vi. When the sample size increases, the standard deviation decreases.
Answer:
3.25
Step-by-step explanation:
In what order should you evaluate problems?
Answer:
(4) → (1) → (3) → (2)
Step-by-step explanation:
Order of operations in any question are decided by the rule,
P → Parentheses
E → Exponents
D → Division
M → Multiplication
A → Addition
S → Subtract
Following the same rule order of operations will be,
- Take care of anything inside the parentheses.
- Evaluate and raise the exponents
- Multiply or divide. Make sure to do whichever one comes first from left to right.
- Add or Subtract from left to right.
Options are arranged in the order of,
(4) → (1) → (3) → (2)
Put the following equation of a line into slope-intercept form, simplifying all
fractions.
3x + 3y = -9
Answer:
[tex]y = -x - 3[/tex]
Step-by-step explanation:
We are trying to get the equation [tex]3x + 3y = -9[/tex] into the form [tex]y = mx+b[/tex], aka slope-intercept form.
To do this we are trying to isolate y.
[tex]3x + 3y = -9[/tex]
Subtract 3x from both sides:
[tex]3y = -9 - 3x[/tex]
Rearrange the terms:
[tex]3y = -3x - 9[/tex]
Divide both sides by 3:
[tex]y = -x - 3[/tex]
Hope this helped!
Grandma baked 969696 cookies and gave them to her grandchildren. One of the grandchildren, Cindy, received ccc fewer cookies than she would have received had all of the cookies been evenly divided among the 888 grandchildren.
Answer:
Answer:
96/8 - c
Step-by-step explanation:
96/8 = 12
The average amount of cookies each grandchild would get is 12. However, Cindy gets c less than the average amount. So, it would be 12 - c.
But, it's asking for the original expression. Therefore, the answer would be 96/8 - c.
Help with number 50 please. Thanks.
Answer:
[tex] d = 7 + 3\sqrt{3} [/tex] and
[tex] d = 7 - 3\sqrt{3} [/tex]
Step-by-step explanation:
To solve the equation, [tex] d^2 - 14d - 22 = 0 [/tex], using the quadratic formula,
Recall: quadratic formula = [tex] \frac{-b ± \sqrt{b^2 - 4ac}}{2a} [/tex]
Where,
a = 1
b = -14
c = 22
Plug in your values into the formula and solve:
[tex] \frac{-(-14) ± \sqrt{(-14)^2 - 4(1)(22)}}{2(1)} [/tex]
[tex] \frac{14 ± \sqrt{196 - 88}}{2} [/tex]
[tex] \frac{14 ± \sqrt{108}}{2} [/tex]
[tex] d = \frac{14 + \sqrt{108}}{2} [/tex]
[tex] d = \frac{14 + 6\sqrt{3}}{2} [/tex]
[tex] d = (\frac{2(7 + 3\sqrt{3})}{2} [/tex]
[tex] d = 7 + 3\sqrt{3} [/tex]
And
[tex] d = \frac{14 - \sqrt{108}}{2} [/tex]
[tex] d = \frac{14 - 6\sqrt{3}}{2} [/tex]
[tex] d = (\frac{2(7 - 3\sqrt{3})}{2} [/tex]
[tex] d = 7 - 3\sqrt{3} [/tex]
Emma rents a car from a company that rents cars by the hour. She has to pay an initial fee of $75, and then they charge her $9 per hour. Write an equation for the total cost if Emma rents the car for ℎ hours. If Emma has budgeted $250 for the rental cars, how many hours can she rent the car? Assume the car cannot be rented for part of an hour.
2/3a - 1/6 =1/3 please help me
Answer:
[tex]a = \frac{3}{4}[/tex]
Step-by-step explanation:
Let's convert everything to sixths to make it easier to work with.
[tex]\frac{4}{6}a - \frac{1}{6} = \frac{2}{6}[/tex]
Add 1/6 to both sides:
[tex]\frac{4}{6}a = \frac{3}{6}[/tex].
Dividing both sides by 4/6:
[tex]a = \frac{3}{6} \div \frac{4}{6}\\\\a = \frac{3}{6} \cdot \frac{6}{4}\\\\a = \frac{18}{24}\\\\a = \frac{3}{4}[/tex]
Hope this helped!
Suppose that $2000 is invested at a rate of 2.6% , compounded semiannually. Assuming that no withdrawals are made, find the total amount after 10 years.
Answer:
$2,589.52
Step-by-step explanation:
[tex] A = P(1 + \dfrac{r}{n})^{nt} [/tex]
We start with the compound interest formula above, where
A = future value
P = principal amount invested
r = annual rate of interest written as a decimal
n = number of times interest is compound per year
t = number of years
For this problem, we have
P = 2000
r = 0.026
n = 2
t = 10,
and we find A.
[tex] A = $2000(1 + \dfrac{0.026}{2})^{2 \times 10} [/tex]
[tex] A = $2589.52 [/tex]
Compound interest formula:
Total = principal x ( 1 + interest rate/compound) ^ (compounds x years)
Total = 2000 x 1+ 0.026/2^20
Total = $2,589.52
. En el triángulo ABC, la medida del ángulo exterior en el vértice B es el triple de la medida del ángulo C y la mediatriz de BC corta a AC en el punto F. sabiendo que FC=12. Calcular AB.
Responder:
| AB | = 12m
Explicación paso a paso:
Verifique el diagrama en el archivo adjunto.
En el diagrama, se puede ver que el lado FC es igual al lado FB de acuerdo con el triángulo isósceles FBC.
Además, el lado FB es igual a AB ya que son paralelos entre sí.
De la declaración anterior, | FC | = | FB | y | FB | = | AB |
Esto significa | FC | = | FB | = | AB |
Por lo tanto desde | FC | = 12 m, | AB | = 12 m ya que ambos lados son iguales.
De ahí el lado | AB | se mide 12m
HELP ASAP
What is the area of the circle shown below?
Answer:
C
Step-by-step explanation:
The area (A) of a circle is calculated as
A = πr² ( r is the radius )
Here r = 18 cm , thus
A = π × 18² = 324π ≈ 1017.9 cm² → C
Answer:
C.) 1017.9 cm²
Step-by-step explanation:
For a given circle
radius (r) = 18 cm
Now,
Area of Circle
= πr²
= 3.14 × (18)² cm
= 3.14 × 324 cm
= 1017.9 cm²
A standardized exam's scores are normally distributed. In a recent year, the mean test score was and the standard deviation was . The test scores of four students selected at random are , , , and . Find the z-scores that correspond to each value and determine whether any of the values are unusual. The z-score for is nothing. (Round to two decimal places as needed.) The z-score for is nothing. (Round to two decimal places as needed.) The z-score for is nothing. (Round to two decimal places as needed.) The z-score for is nothing. (Round to two decimal places as needed.) Which values, if any, are unusual? Select the correct choice below and, if necessary, fill in the answer box within your choice. A. The unusual value(s) is/are nothing. (Use a comma to separate answers as needed.) B. None of the values are unusual.
Answer:
The z-score for 1880 is 1.34.
The z-score for 1190 is -0.88.
The z-score for 2130 is 2.15.
The z-score for 1350 is -0.37.
And the z-score of 2130 is considered to be unusual.
Step-by-step explanation:
The complete question is: A standardized exam's scores are normally distributed. In recent years, the mean test score was 1464 and the standard deviation was 310. The test scores of four students selected at random are 1880, 1190, 2130, and 1350. Find the z-scores that correspond to each value and determine whether any of the values are unusual. The z-score for 1880 is nothing. (Round to two decimal places as needed.) The z-score for 1190 is nothing. (Round to two decimal places as needed.) The z-score for 2130 is nothing. (Round to two decimal places as needed.) The z-score for 1350 is nothing. (Round to two decimal places as needed.) Which values, if any, are unusual? Select the correct choice below and, if necessary, fill in the answer box within your choice. A. The unusual value(s) is/are nothing. (Use a comma to separate answers as needed.) B. None of the values are unusual.
We are given that the mean test score was 1464 and the standard deviation was 310.
Let X = standardized exam's scores
The z-score probability distribution for the normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean test score = 1464
[tex]\sigma[/tex] = standard deviation = 310
S, X ~ Normal([tex]\mu=1464, \sigma^{2} = 310^{2}[/tex])
Now, the test scores of four students selected at random are 1880, 1190, 2130, and 1350.
So, the z-score of 1880 = [tex]\frac{X-\mu}{\sigma}[/tex]
= [tex]\frac{1880-1464}{310}[/tex] = 1.34
The z-score of 1190 = [tex]\frac{X-\mu}{\sigma}[/tex]
= [tex]\frac{1190-1464}{310}[/tex] = -0.88
The z-score of 2130 = [tex]\frac{X-\mu}{\sigma}[/tex]
= [tex]\frac{2130-1464}{310}[/tex] = 2.15
The z-score of 1350 = [tex]\frac{X-\mu}{\sigma}[/tex]
= [tex]\frac{1350-1464}{310}[/tex] = -0.37
Now, the values whose z-score is less than -1.96 or higher than 1.96 are considered to be unusual.
According to our z-scores, only the z-score of 2130 is considered to be unusual as all other z-scores lie within the range of -1.96 and 1.96.
The blood platelet counts of a group of women have a bell-shaped distribution with a mean of 256.3 and a standard deviation of 66.8. (All units are 1000 cells/μL.) Using the empirical rule, find each approximate percentage below. a. What is the approximate percentage of women with platelet counts within 3 standard deviations of the mean, or between 55.9 and 456.7? b. What is the approximate percentage of women with platelet counts between 122.7 and 389.9?
Answer:
a) In the interval ( 55,9 ; 456,7 ) we will find 99,7 % of all values
b) In the interval ( 122,7 ; 389,9 ) we find 95,4 % of all values
Step-by-step explanation:
For a Normal distribution N (μ ; σ ) the Empirical rule establishes that the intervals:
( μ ± σ ) contains 68,3 % of all values
( μ ± 2σ ) contains 95,4 % of all values
( μ ± 3σ ) contains 99,7 % of all values
If N ( 256,3 ; 66,8 )
σ = 66,8 ⇒ 3*σ = 3 * 66,8 = 200,4
Then: 256,3 - 200,4 = 55,9
And 256,3 + 200,4 = 456,7
a) In the interval ( 55,9 ; 456,7 ) we will find 99,7 % of all values
b) 2*σ = 2 * 66,8 = 133,6
Then 256,3 - 133,6 = 122,7
And 256,3 + 133,6 = 389,90
Then in the interval ( 122,7 ; 389,9 ) we find 95,4 % of all values
Please help. I’ll mark you as brainliest if correct.
Answer:
Infinite number of solutions.
Step-by-step explanation:
There are an infinite number of solutions. If you graph both lines, you find they are the same line. If you multiply the send equation by -4, you’ll end up with the first equation. I’m not sure what your teacher means by specifying their form.
Question 15
FLAG QUESTION
You bought 9.5 pounds of chicken for $12.73. Find the cost of one pound of chicken.
Answer:
One pound of chicken is $1.34
Step-by-step explanation:
So, if 9.5 pounds of chicken is $12.73, all you have to do it divide 12.73 and 9.5. That way, you can see how much each pound is separately! Hopefully this helps!
Answer:
[tex]\huge\boxed{1\ pound\ of\ chicken = \$1.34}[/tex]
Step-by-step explanation:
9.5 pounds of chicken = $12.73
Dividing both sides by 9.5
1 pound of chicken = $12.73 / 9.5
1 pound of chicken = $1.34
The numbers of words defined on randomly selected pages from a dictionary are shown below. Find the mean, median, mode of the listed numbers. 30 31 64 59 57 33 54 77 56 41 What is the mean? Select the correct choice below and ,if necessary ,fill in the answer box within your choice.(around to one decimal place as needed)
Answer:
mean=502/10=50.2
median=(54+56)2=55
Solve the following system of equations using the elimination method. x – y = 11 2x + y = 19
━━━━━━━☆☆━━━━━━━
▹ Answer
(10, -1)
▹ Step-by-Step Explanation
x - y = 11
2x + y = 19
Sum up the equations:
3x = 30
Divide 3 on both sides:
x = 10
Substitute:
10 - y = 11
y = -1
Solution:
(x, y) (10, -1)
Hope this helps!
CloutAnswers ❁
━━━━━━━☆☆━━━━━━━
In a genetics experiment on peas, one sample of offspring contained green peas and yellow peas. Based on those results, estimate the probability of getting an offspring pea that is green. Is the result reasonably close to the value of that was expected? 350 127 3 4 The probability of getting a green pea is approximately . (Type an integer or decimal rounded to three decimal places as needed.) Is this probability reasonably close to ? Choose the correct answer below. 3 4 A. No, it is not reasonably close. B. Yes, it is reasonably close.
Answer:
The probability of getting an offspring pea that is green is is 0.733
YES, the probability is reasonably close to the expected value of 3/4 (0.750)
Step-by-step explanation:
The formula for calculating the probability of an event is;
P = Favorable Outcome / Sample space
Let A be an event of getting an offspring green peas, B be an event of getting an offspring yellow peas and N be the total number of peas.
number of green peas in an offspring are 350
number of yellow peas in an offspring are 127
total number of peas are 477
So in the genetic experiment, the number of times event A occurs is 350 and the number times event B occurs is 127
Now the probability of getting an offspring pea that is green is
P = number of green peas / total number of peas
p = n(A)/N
p = 350/477
p = 0.733
So YES, the probability is reasonably close to 3/4 ( 0.750 )
The probability of getting an offspring pea that is green is is 0.733.
YES, the probability is reasonably close to the expected value of 3/4 (0.750)
The U.S. Dairy Industry wants to estimate the mean yearly milk consumption. A sample of 21 people reveals the mean yearly consumption to be 74 gallons with a standard deviation of 16 gallons. Assume that the population distribution is normal. (Use t Distribution Table.)
a-1. What is the value of the population mean?
16
Unknown
74
a-2. What is the best estimate of this value?
Estimate population mean
c. For a 90% confidence interval, what is the value of t? (Round your answer to 3 decimal places.)
Value of t
d. Develop the 90% confidence interval for the population mean. (Round your answers to 3 decimal places.)
Confidence interval for the population mean is and .
e. Would it be reasonable to conclude that the population mean is 68 gallons?
a) Yes
b) No
c) It is not possible to tell.
Correct question is;
The U.S. Dairy Industry wants to estimate the mean yearly milk consumption. A sample of 21 people reveals the mean yearly consumption to be 74 gallons with a standard deviation of 16 gallons.
a. What is the value of the population mean? What is the best estimate of this value?
b. Explain why we need to use the t distribution. What assumption do you need to make?
c. For a 90 percent confidence interval, what is the value of t?
d. Develop the 90 percent confidence interval for the population mean.
e. Would it be reasonable to conclude that the population mean is 68 gallons?
Answer:
A) Best estimate = 74 gallons
B) because the population standard deviation is unknown. The assumption we will make is that the population follows the normal distribution.
C) t = 1.725
D) 90% confidence interval for the population mean is (67.9772, 80.0228) gallons
E) Yes
Step-by-step explanation:
We are given;
Sample mean; x' = 74
Sample population; n = 21
Yearly Standard deviation; s = 16
A) We are not given the population mean.
So the closest estimate to the population mean would be the sample mean which is 74.
B) We are not given the population standard deviation and as such we can't use normal distribution. So what is used when population standard deviation is not known is called t - distribution table. The assumption we will make is that the population follows the normal distribution.
C) At confidence interval of 90% and DF = n - 1 = 21 - 1 = 20
From t-tables, the t = 1.725
D) Formula for the confidence interval is;
x' ± t(s/√n) = 74 ± 1.725(16/√21) = 74 ± 6.0228 = 67.9772 or 80.0228
Thus 90% confidence interval for the population mean is (67.9772, 80.0228) gallons
E) 68 gallons lies within the range of the confidence interval, thus we can say that "Yes, it is reasonable"
A regression was run to determine if there is a relationship between hours of TV watched per day (x) and the number of sit-ups a person can do (y). The results were: y = a+bx b = -0.89 a = 23.65 r2 = 0.7038 If a person watches 14 hours of television a day, predict how many sit-ups he can do. What is the value of the correlation coefficient? Round to three decimal places.
Answer:
y = 11.19 ; 0.839
Step-by-step explanation:
Given the following :
relationship between hours of TV watched per day (x) and the number of sit-ups a person can do (y)
y = a + bx ; comparing with the linear regression model function
y = predicted variable
a = intercept
b = slope or gradient
x = independent variable
b = -0.89 a = 23.65 r2 = 0.7038
Therefore, if a person watches for 14 hours per day, that is x = 14, the number of sit-ups he can do will be :
y = 23.65 + (-0.89)(14)
y = 23.65 - 12.46
y = 11.19
About 11 sit-ups.
If the r^2 value = 0.7038
Then the Coefficient of regression = r
Will be the square root of r^2
r = sqrt(r^2)
r = sqrt(0.7038)
r =0.8389278 = 0.839
Suppose that 11% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts, and let X denote the number among these that are nonconforming and can be reworked.Required:a. What is the (approximate) probability that X is at most 30?b. What is the (approximate) probability that X is less than 30?c. What is the (approximate) probability that X is between 15 and 25 (inclusive)?
Answer:
(a) The probability that X is at most 30 is 0.9726.
(b) The probability that X is less than 30 is 0.9554.
(c) The probability that X is between 15 and 25 (inclusive) is 0.7406.
Step-by-step explanation:
We are given that 11% of all steel shafts produced by a certain process are nonconforming but can be reworked. A random sample of 200 shafts is taken.
Let X = the number among these that are nonconforming and can be reworked
The above situation can be represented through binomial distribution such that X ~ Binom(n = 200, p = 0.11).
Here the probability of success is 11% that this much % of all steel shafts produced by a certain process are nonconforming but can be reworked.
Now, here to calculate the probability we will use normal approximation because the sample size if very large(i.e. greater than 30).
So, the new mean of X, [tex]\mu[/tex] = [tex]n \times p[/tex] = [tex]200 \times 0.11[/tex] = 22
and the new standard deviation of X, [tex]\sigma[/tex] = [tex]\sqrt{n \times p \times (1-p)}[/tex]
= [tex]\sqrt{200 \times 0.11 \times (1-0.11)}[/tex]
= 4.42
So, X ~ Normal([tex]\mu =22, \sigma^{2} = 4.42^{2}[/tex])
(a) The probability that X is at most 30 is given by = P(X < 30.5) {using continuity correction}
P(X < 30.5) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{30.5-22}{4.42}[/tex] ) = P(Z < 1.92) = 0.9726
The above probability is calculated by looking at the value of x = 1.92 in the z table which has an area of 0.9726.
(b) The probability that X is less than 30 is given by = P(X [tex]\leq[/tex] 29.5) {using continuity correction}
P(X [tex]\leq[/tex] 29.5) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{29.5-22}{4.42}[/tex] ) = P(Z [tex]\leq[/tex] 1.70) = 0.9554
The above probability is calculated by looking at the value of x = 1.70 in the z table which has an area of 0.9554.
(c) The probability that X is between 15 and 25 (inclusive) is given by = P(15 [tex]\leq[/tex] X [tex]\leq[/tex] 25) = P(X < 25.5) - P(X [tex]\leq[/tex] 14.5) {using continuity correction}
P(X < 25.5) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{25.5-22}{4.42}[/tex] ) = P(Z < 0.79) = 0.7852
P(X [tex]\leq[/tex] 14.5) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{14.5-22}{4.42}[/tex] ) = P(Z [tex]\leq[/tex] -1.70) = 1 - P(Z < 1.70)
= 1 - 0.9554 = 0.0446
The above probability is calculated by looking at the value of x = 0.79 and x = 1.70 in the z table which has an area of 0.7852 and 0.9554.
Therefore, P(15 [tex]\leq[/tex] X [tex]\leq[/tex] 25) = 0.7852 - 0.0446 = 0.7406.
Show the distributive property can be used to evaluate 7x8 4/5
Answer:
308/5
Step-by-step explanation:
It can be Written as
= 7 ×(8 + 4/5 )
= 7 × 8 + 7 × 4/5
= 56 + 28/5
= 280+28
________
5
= 308/5
How do you write 5.44 in words?
Answer:
five and forty-four hundredths
Step-by-step explanation:
Answer:
five point four four
Step-by-step explanation:
