Final answer:
To find the fraction of applicants with a score of 400 or above, convert the score to a z-score and look it up in a standard normal distribution table. Subtract the resulting proportion from 1 to find the fraction above 400. Approximately 77.34% of the applicants would have a score of 400 or above.
Explanation:
To find the fraction of applicants who would have a score of 400 or above, we need to find the area under the normal distribution curve to the right of 400. First, we need to convert the score of 400 to a z-score using the formula:
z = (x - μ) / σ
where z is the z-score, x is the score, μ is the mean, and σ is the standard deviation. In this case, the mean is 460 and the standard deviation is 80, so the z-score is:
z = (400 - 460) / 80 = -0.75
Once we have the z-score, we can look it up in a standard normal distribution table to find the proportion of the distribution that is below it. The table gives us a value of approximately 0.2266 for a z-score of -0.75. Since we want the fraction above 400, we can subtract this value from 1 to get:
1 - 0.2266 = 0.7734
Therefore, we would expect approximately 77.34% of the applicants to have a score of 400 or above.
A random sample of 20 students yielded a mean of ¯x = 72 and a variance of s2 = 16 for scores on a college placement test in mathematics. Assuming the scores to be normally distributed, construct a 98% confidence interval for σ2.
Answer:
The 98% confidence interval for the variance in the pounds of impurities would be [tex]8.400 \leq \sigma^2 \leq 39.827[/tex].
Step-by-step explanation:
1) Data given and notation
[tex]s^2 =16[/tex] represent the sample variance
s=4 represent the sample standard deviation
[tex]\bar x[/tex] represent the sample mean
n=20 the sample size
Confidence=98% or 0.98
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 .
2) Calculating the confidence interval
The confidence interval for the population variance is given by the following formula:
[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]
On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.
The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:
[tex]df=n-1=20-1=19[/tex]
Since the Confidence is 0.98 or 98%, the value of [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], and we can use excel, a calculator or a table to find the critical values.
The excel commands would be: "=CHISQ.INV(0.01,19)" "=CHISQ.INV(0.99,19)". so for this case the critical values are:
[tex]\chi^2_{\alpha/2}=36.191[/tex]
[tex]\chi^2_{1- \alpha/2}=7.633[/tex]
And replacing into the formula for the interval we got:
[tex]\frac{(19)(16)}{36.191} \leq \sigma^2 \leq \frac{(19)(16)}{7.633}[/tex]
[tex] 8.400 \leq \sigma^2 \leq 39.827[/tex]
So the 98% confidence interval for the variance in the pounds of impurities would be [tex]8.400 \leq \sigma^2 \leq 39.827[/tex].
The 98% confidence interval for the mean change in score is between 69.92 points to 74.08 points
How to calculate confidence intervalStandard deviation = √variance = √16 = 4
The z score of 98% confidence interval is 2.326
The margin of error (E) is:
[tex]E = Z_\frac{\alpha }{2} *\frac{standard\ deviation}{\sqrt{sample\ size} } =2.326*\frac{4}{\sqrt{20} } =2.08[/tex]
The confidence interval = mean ± E = 72 ± 2.08 = (69.92, 74.08)
The 98% confidence interval for the mean change in score is between 69.92 points to 74.08 points
Find out more on confidence interval at: https://brainly.com/question/15712887
You have a fair coin. You flip the coin two times. Let T1T1 be the event that the first flip (Flip 1) results in Tails. Let T2T2 be the event that the second flip (Flip 2) results in Tails. Are the events T1T1 and T2T2 independent?
Answer:
Yes, the events T1 and T2 are independent.
Step-by-step explanation:
When flipping a fair coin, for each flip, there is a 50/50 chance that it will result in heads or tails.
For the first flip, P(T1) = 0.5
For the second flip, P(T2) = 0.5 regardless of the outcome of the first flip. Therefore, T1 and T2 are independent events.
*Note that if the question asked for the event of both flips resulting in tails, then the events would be dependent.
A researcher developing scanners to search for hidden weapons at airports has concluded that a new device is significantly better than the current scanner. He made this decision based on a test using a = 0.05. Would he have made the same decision at a = 0.10? How about a = 0.01? Explain.
Answer:
He can make the same decision at a = 0.10, but he may not make the same decision at a = 0.01
Step-by-step explanation:
In hypothesis tests, critical regions are ranges of the distributions where the values represent statistically significant results. Analysts define the size and location of the critical regions by specifying both the significance level (alpha) and whether the test is one-tailed or two-tailed.
As the significance level gets bigger, the range of the critical region increases.
Therefore significant results at lower significance levels are still significant at higher significance levels. Thus significant result at a=0.05 is always significant at a=0.10
But significant result at a=0.05 may not be significant at a=0.01 since critical region shrinks, therefore the result may not fall in the critical region at a=0.01
Consider a binomial experiment with n = 10 and p = 0.10. (a) Compute f(0). If required, round your answer to four decimal places. (b) Compute f(2). If required, round your answer to four decimal places. (c) Compute P(x ≤ 2). If required, round your answer to four decimal places. (d) Compute P(x ≥ 1). If required, round your answer to four decimal places. (e) Compute E(x). (f) Compute Var(x) and σ. If required, round Var(x) answer to one decimal place and σ answer to four decimal places. Var(x) = σ =
Answer:
Step-by-step explanation:
Hello!
You have X~Bi (n;ρ)
Where:
n=10
ρ= 0.10
For all asked probabilities you need to use a Binomial distribution table. Remember this table has the information of the cummulative probabilities P(X≤x).
a. f(0) ⇒ P(X=0) = 0.3487
b. f(2) ⇒ P(X=2) ⇒ P(X≤2) - P(X≤1) = 0.9298 - 0.7361 = 0.1937
c. P(X≤2) = 0.9298
d. P(X ≥ 1) = 1 - P(X ≤ 1) = 1 - 0.7361 = 0.2639
e. E(X)= nρ = 10*0.10 = 1
f. V(X)= nρ(1-ρ) = 10*0.1*0.9 = 0.9
σ= √V(X) = √0.9 = 0.9487
I hope it helps!
The binomial experiment depicts that f(0) will be 0.3487.
How to compute the binomial experimentFrom the information given, it can be noted that:
n = 10
p = 0.10
q = 1 - 0.10 = 0.90
f(0) ⇒ P(X=0) = 0.3487
f(2) = P(X=2) ⇒ P(X≤2) - P(X≤1)
= 0.9298 - 0.7361
= 0.1937
P(X≤2) = 0.9298
P(X ≥ 1) = 1 - P(X ≤ 1)
= 1 - 0.7361 = 0.2639
E(X) = nρ
= 10 × 0.10 = 1
V(X) = nρ(1-ρ)
= 10 × 0.1 × 0.9
= 0.9
σ = √V(X) = √0.9
= 0.9487
Learn more about binomial on:
https://brainly.com/question/2284337
Bob and Sally mortgage payment increases to $1632.00. Their roof starts to leak and they need to replace it. They budget 500$ per month (replace the old student loan charge with roof expense).
Answer:
...
Step-by-step explanation:
Which system of linear equations is graphed below
Answer:
D
Step-by-step explanation:
what are the next two terms in the sequence -16, 4, -1, 1/4
Answer: fifth term = -1/16
Sixth term = 1/256
Step-by-step explanation:
The given sequence is a geometric progression. This is because the ratio of two consecutive terms is constant.
We will apply the formula for determining the nth term of a geometric progression series.
Tn = ar^n-1
Where
Tn = value of the nth term of the geometric series
a = The first term of the series.
r = common ratio(ratio of a term to a consecutive previous term)
n = number of terms in the series
From the in information given,
a = -16
r = 4/-16 = -1/4
The next 2 terms are the 5th and 6th terms.
T5 = -16 × -1/4^(5-1)
T5 = -16 × (-1/4)^4
T5 = -16 × 1/256 = -1/16
The 6th term would be the 5th term × the common ratio. It becomes
T6 = -1/16 ×-1/4 = 1/256
. Among all the income tax forms filed in a certain year, the mean tax paid was $2000, and the standard deviation was $500. In addition, for 10% of the forms, the tax paid was greater than $3000. A random sample of 625 tax forms is drawn. a. What is the probability that the average tax paid on the sample forms is greater than $1980? b. What is the probability that more than 60 of the sampled forms have a tax of greate
To find the probability that the average tax paid on the sample forms is greater than $1980, standardize the sample mean using the Central Limit Theorem. To find the probability that more than 60 of the sampled forms have a tax greater than $3200, use a normal distribution approximation.
Explanation:To solve this problem, we can use the Central Limit Theorem.
a. To find the probability that the average tax paid on the sample forms is greater than $1980, we need to standardize the sample mean. We calculate the z-score by subtracting the population mean from the sample mean and dividing by the population standard deviation divided by the square root of the sample size. Then, we can use a z-table or a calculator to find the probability.
b. To find the probability that more than 60 of the sampled forms have a tax greater than $3200, we can use a normal distribution approximation. We can calculate the z-score for 60 forms by subtracting the population mean from 60 and dividing by the square root of the population mean multiplied by (1 - the population mean) divided by the sample size. Then, we can use a z-table or a calculator to find the probability.
Learn more about Probability here:https://brainly.com/question/32117953
#SPJ6
A truck driver operates a delivery service in a southern city. His start-up costs amounted to $2500. He estimates that it costs him (in terms of gasoline, wear and tear on his truck, etc.)$3.00 per delivery. He charges $5.50 per delivery. Let x represent the number of deliveries he makes. (a) Express the cost C as a function of x. (b) Express the revenue R as a function of x. (c) Determine analytically the value of x for which revenue equals cost. (d) Graph y1equalsC(x) and y2equalsR(x) on the same xy-axes and interpret the graphs.
Answer:
(a)C(x) = 2500 + 3x
(b)R(x) = 5.5x
(c)x = 1000
Step-by-step explanation:
(a)His cost function as a function of x
C(x) = 2500 + 3x
(b)His revenue R function as a function of x
R(x) = 5.5x
(c)When revenue R equals to Cost C
C(x) = R(x)
2500 + 3x = 5.5x
2.5x = 2500
x = 1000
(d) Refer to attachment. We can see that the 2 lines intercepts at x = 1000 and y = 2500. That's the point of turning to profit.
The product of an initial investment, I, and the quantity of the sum of 1 and the annual interest rate, r, raised to the power of n, the number of years of the investment, is equal to M, the current amount of money in an investment account. If an initial investment of $1,423.00 is made to an account with an annual interest rate of 2%, what will be the value of M after 3 years? Round to the nearest cent. A. $1,510.10 B. $1,339.32 C. $1,480.49 D. $1,508.38
Answer:
A. $1,510.10
Step-by-step explanation:
Fill in the given numbers and do the arithmetic.
M = I·(1+r)^n
M = $1423.00·(1 +.02)^3 = $1510.099
M ≈ $1510.10
Suppose that we are testing a coin to see if it is fair, so our hypotheses are: H0: p = 0.5 vs Ha: p ≠ 0.5. In each of (a) and (b) below, use the "Edit Data" option on StatKey to find the p-value for the sample results and give a conclusion in the test. a. We get 56 heads out of 100 tosses. b. We get 560 heads out of 1000 tosses. c. Compare the sample proportions in parts (a) and (b). Compare the p-values. Why are the p-values so different?
For testing if a coin is fair, a binomial test is used to compare the observed heads to what's expected under a fair coin scenario. Differences in p-values for the same sample proportion across different sample sizes are due to the increased precision of larger samples.
In hypothesis testing, when we want to determine whether a coin (or in this case, a hypothetical lizard) is fair, we employ a binomial test. This test assesses the number of 'successes' (heads in our case), comparing it to what we would expect under the null hypothesis of p=0.5, indicating a fair coin.
For part (a), where we get 56 heads out of 100 tosses, we would calculate the p-value by looking at both the proportion of heads obtained and the standard error for a binomial distribution. Although not calculated here directly, if this p-value is less than the chosen significance level, typically 0.05, we reject the null hypothesis, concluding the coin is not fair.
For part (b), getting 560 heads out of 1000 tosses yields the same sample proportion as part (a). However, the larger sample size increases the power of the test, often resulting in a smaller p-value if the observed proportion is different from the expected 0.5.
Comparing p-values from (a) and (b) reveals that despite having the same sample proportions, the p-values differ due to the sample sizes. Larger sample sizes yield more precise estimates of the population proportion, thus potentially resulting in smaller p-values for the same sample proportion deviation from the null hypothesis.
A toy company is modeling a real home as a doll house. They decide to use a 2/3 inch to 1 foot scale, the traditional scale for doll house built in the 20th century. If the actual house is 30 feet high, how high will the dollhouse be in inches?
Answer:
20 inches
Step-by-step explanation:
Take 2/3 of an inch 30 times.
The height of the model is then 30 times 2/3 inches = 10*2 = 20 inches
Which of the following is the purpose of a confidence interval for the fitted (predicted value) from a
regression model?
A) To estimate the value of a future observation for a given value of x.
B) To estimate the population mean of Y for a given value of x
C) To estimate the population slope of the regression model
D) To estimate the sample mean of Y for a given value of x.
Answer: option A is correct
Step-by-step explanation:
the prediction interval is to cover a “moving target”, the random future value of y,
while the confidence interval is to cover the “fixed target”,
the average (expected) value of y, E(y).
Willow Brook National Bank operates a drive-up teller window that allows customers to complete bank transactions without getting out of their cars. On weekday mornings, arrivals to the drive-up teller window occur at random, with an arrival rate of 24 customers per hour or 0.4 customers per minute.
(a) What is the mean or expected number of customers that will arrive in a five-minute period?
(b) Use the arrival rate in part (a) and compute the probabilities that exactly 0, 1, 2, and 3 customers will arrive during a five-minute period. If required, round your answers to four decimal places.
(c) Delays are expected if more than three customers arrive during any five-minute period. What is the probability that delays will occur? If required, round your answer to four decimal places.
Answer:
2, 0.135, 0.270, 0.270, 0.180, 0.145
Step-by-step explanation:
1. To calculate the mean of people arriving in 5 minute period:
We know the arrival rate of minute is 0.4, so people arriving in 5 minutes will be
0.4 x 5 = 2
2. From part 1 it is known thay mean arrival time=2
For this we will use poisson's probability formula that is
P(X=x) = (2^x) x Exp^(-2/x!)
For X=0
P(X=0) = (2^0) x Exp^(-2/0!) = 0.135
For X = 1
P(X=1) = (2^1) x Exp^(-2/1!) = 0.270
For X = 2
P(X=2) = (2^2) x Exp^(-2/2!) = 0.270
For X = 3
P(X=3) = (2^3) x Exp^(-2/3!) = 0.180
3. For delay expected if more than 3 customer arrive in 5 minutes.
P(X>3) = 1 - P(X=0) - P(X=1) - P(X=2) - P(X=3)
P(X>3) = 1-0.135-0.270-0.270-0.180
P(X>3) = 0.145
A supervisor records the repair cost for 25 randomly selected dryers. A sample mean of $93.36 and standard deviation of $19.95 are subsequently computed. Determine the 98% confidence interval for the mean repair cost for the dryers. Assume the population is approximately normal.
Answer:
The 98% confidence interval for the mean repair cost for the dryers is (83.4161, 103.3039).
Step-by-step explanation:
We have a small sample size n = 25, [tex]\bar{x} = 93.36[/tex] and s = 19.95. The confidence interval is given by [tex]\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})[/tex] where [tex]t_{\alpha/2}[/tex] is the [tex]\alpha/2[/tex]th quantile of the t distribution with n - 1 = 25 - 1 = 24 degrees of freedom. As we want the 98% confidence interval, we have that [tex]\alpha = 0.02[/tex] and the confidence interval is [tex]93.36\pm t_{0.01}(\frac{19.95}{\sqrt{25}})[/tex] where [tex]t_{0.01}[/tex] is the 1st quantile of the t distribution with 24 df, i.e., [tex]t_{0.01} = -2.4922[/tex]. Then, we have [tex]93.36\pm (-2.4922)(\frac{19.95}{\sqrt{25}})[/tex] and the 98% confidence interval is given by (83.4161, 103.3039).
Answer:
Step-by-step explanation:
Our aim is to determine a 98% confidence interval for the mean repair cost for the dryers
Number of samples. n = 25
Mean, u = $93.36
Standard deviation, s = $19.95
For a confidence level of 98%, the corresponding z value is 2.33. This is determined from the normal distribution table.
We will apply the formula,
Confidence interval
= mean ± z × standard deviation/√n
It becomes
93.36 ± 2.33 × 19.95/√25
= 93.36 ± 2.33 × 3.99
= 93.36 ± 9.2967
The lower boundary of the confidence interval is 93.36 - 9.2967 =84.0633
The upper boundary of the confidence interval is 93.36 + 9.2967 = 102.6567
Therefore, with 98% confidence interval, the mean repair costs for the dryers is between $84.0633 and $102.6567
A typist makes an average of 5 typo errors per page that she types. If the number of errors per each typed page can be modeled according to the Poisson distribution, what is the probability that she will make more than 3 typo errors in the next page? Express your answer in decimals and round to two significant places after the decimal. For example 0.1678 should be entered as 0.17.
Answer:
Step-by-step explanation:
Marie sees a leather jacket kn sale for $73.00. The sign says 20% off. How much was the jacket originally?
Answer:
It was originally $91.25
The cycle time for trucks hauling concrete to a highway construction site is uniformly distributed over the interval 50 to 70 minutes. What is the conditional probability that the cycle time is no more than 65 minutes if it is known that the cycle time exceeds 55 minutes?
Answer:
2/3=0.6667
Step-by-step explanation:
Let X be the cycle time for trucks hauling concrete to a highway construction site
Given that X is U(50,70)
Hence pdf of X is
[tex]f(x) = \frac{1}{20} ,50<x<70[/tex]
Let A be the event that cycle time is no more than 65 minutes and
B the event cycle time exceeds 55 minutes
Required probability
= the conditional probability that the cycle time is no more than 65 minutes if it is known that the cycle time exceeds 55 minutes
= P(A/B)
=[tex]\frac{P(A\bigcapB)}{P(B)} \\=\frac{P(55<x<65)}{P(X>55)} \\=\frac{65-55}{70-55} \\=\frac{2}{3}[/tex]
A fair coin is tossed three times and the events A, B, and C are defined as follows: A:{ At least one head is observed } B:{ At least two heads are observed } C:{ The number of heads observed is odd } Find the following probabilities by summing the probabilities of the appropriate sample points (note that 0 is an even number):
(a) P(B) =
(b) P(A or B) =
(c) P(A or B or C)
Answer:
(a) 1/2
(b) 1/2
(c) 1/8
Step-by-step explanation:
Since, when a fair coin is tossed three times,
The the total number of possible outcomes
n(S) = 2 × 2 × 2
= 8 { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT },
Here, B : { At least two heads are observed } ,
⇒ B = {HHH, HHT, HTH, THH},
⇒ n(B) = 4,
Since,
[tex]\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}[/tex]
(a) So, the probability of B,
[tex]P(B) =\frac{n(B)}{n(S)}=\frac{4}{8}=\frac{1}{2}[/tex]
(b) A : { At least one head is observed },
⇒ A = {HHH, HHT, HTH, THH, HTT, THT, TTH},
∵ A ∩ B = {HHH, HHT, HTH, THH},
n(A∩ B) = 4,
[tex]\implies P(A\cap B) = \frac{n(A\cap B)}{n(S)} = \frac{4}{8}=\frac{1}{2}[/tex]
(c) C: { The number of heads observed is odd },
⇒ C = { HHH, HTT, THT, TTH},
∵ A ∩ B ∩ C = {HHH},
⇒ n(A ∩ B ∩ C) = 1,
[tex]\implies P(A\cap B\cap C)=\frac{1}{8}[/tex]
Young's modulus is a quantitative measure of stiffness of an elastic material. Suppose that for metal sheets of a particular type, its mean value and standard deviation are 70 GPa and 2.2 GPa, respectively. Suppose the distribution is normal. (Round your answers to four decimal places.) (a) Calculate P(69 ≤ X ≤ 71) when n = 16.
The P(69 ≤ X ≤ 71) when n = 16 is approximately 0.9312.
Here's the calculation for P(69 ≤ X ≤ 71) when n = 16:
1. Find the standard error (SE):
SE = σ / √n = 2.2 GPa / √16 = 0.55 GPa
2. Standardize the values:
Z1 = (69 - 70) / 0.55 = -1.82
Z2 = (71 - 70) / 0.55 = 1.82
3. Use a standard normal table or calculator to find the probabilities:
P(Z ≤ 1.82) = 0.9656
P(Z ≤ -1.82) = 0.0344
4. Calculate the probability of the interval:
P(69 ≤ X ≤ 71) = P(-1.82 ≤ Z ≤ 1.82) = P(Z ≤ 1.82) - P(Z ≤ -1.82)
= 0.9656 - 0.0344 = 0.9312
Therefore, P(69 ≤ X ≤ 71) when n = 16 is approximately 0.9312.
A food safety guideline is that the mercury in fish should be below 1 part per million? (ppm). Listed below are the amounts of mercury? (ppm) found in tuna sushi sampled at different stores in a major city. Construct a 99?% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna? sushi?
a. 0.59
b. 0.68
c. 0.10
d. 0.95
e. 1.23
f. 0.59
g. 0.92
Answer:
can you provide the list?
Step-by-step explanation:
The average life a manufacturer's blender is 5 years, with a standard deviation of 1 year. Assuming that the lives of these blenders follow approximately a normal distribution, find the probability that the mean life a random sample of 25 such blenders falls between 4.6 and 5.1 years.
Answer:
The probability that the mean life a random sample of 25 such blenders falls between 4.6 and 5.1 years is 0.6687.
Step-by-step explanation:
We have given :
The average life a manufacturer's blender is 5 years i.e. [tex]\mu=5[/tex]
The standard deviation is [tex]\sigma=1[/tex]
Number of sample n=25.
To find : The probability that the mean life falls between 4.6 and 5.1 years ?
Solution :
Using z-score formula, [tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
The probability that the mean life falls between 4.6 and 5.1 years is given by, [tex]P(4.6<X<5.1)[/tex]
[tex]P(4.6<X<5.1)=P(\frac{4.6-5}{\frac{1}{\sqrt{25}}}<Z<\frac{5.1-5}{\frac{1}{\sqrt{25}}})[/tex]
[tex]P(4.6<X<5.1)=P(\frac{-0.4}{\frac{1}{5}}<Z<\frac{0.1}{\frac{1}{5}})[/tex]
[tex]P(4.6<X<5.1)=P(-2<Z<0.5)[/tex]
[tex]P(4.6<X<5.1)=P(Z<0.5)-P(Z<-2)[/tex]
Using z-table,
[tex]P(4.6<X<5.1)=0.6915-0.0228[/tex]
[tex]P(4.6<X<5.1)=0.6687[/tex]
Therefore, the probability that the mean life a random sample of 25 such blenders falls between 4.6 and 5.1 years is 0.6687.
The probability that the mean life of a random sample of 25 blenders falls between 4.6 and 5.1 years is approximately 0.3585, or 35.85%.
Explanation:To find the probability that the mean life of a random sample of 25 blenders falls between 4.6 and 5.1 years, we can use the standard normal distribution. First, we need to standardize the values using the z-score formula. The z-score for 4.6 years is (4.6 - 5) / 1 = -0.4, and the z-score for 5.1 years is (5.1 - 5) / 1 = 0.1. We can then look up the corresponding probabilities in the standard normal distribution table or use a calculator to find the area under the curve between these two z-scores. The probability that the mean life falls in this range is approximately 0.3585, or 35.85%.
Learn more about Probability here:https://brainly.com/question/32117953
#SPJ11
given f(x)= -x^ {2 }+10x-3, find f(-1)
Answer: f(-) = - 12
Step-by-step explanation:
The function is expressed as
f(x)= -x^2+10x-3
To determine f(-1), we will substitute
x = - 1 into the given function. It becomes
f(-1) = (- 1)^2 + 10(-1) - 3
f(-1) = 1 - 10 - 3
f(-1) = - 12
A random sample of size 100 was taken from a population. A 94% confidence interval to estimate the mean of the population was computed based on the sample data. The confidence interval for the mean is: (107.62, 129.75). What is the z-value that was used in the computation. Round your z-value to 2 decimal places. Select one: A. 1.45 B. 1.55 C. 1.73 D. 1.88 E. 1.96 E-mail fraud (phishing) is becoming an increasing problem for users of the internet. Suppose that 70% of all internet users experience e-mail fraud. If 50 internet users were randomly selected, what is the probability that no more than 25 were victims of e-mail fraud? Answer:
Answer:
Step-by-step explanation:
1) The z value was determined using a normal distribution table. From the normal distribution table, the corresponding z value for a 94% confidence interval is 1.88
The correct option is D
2) if 70% of all internet users experience e-mail fraud. It means that probability of success, p
p = 70/100 = 0.7
q = 1 - p = 1 - 0.7 = 0.3
n = number of selected users = 50
Mean, u = np = 50×0.7 = 35
Standard deviation, u = √npq = √50×0.7×0.3 = 3.24
x = number of internet users
The formula for normal distribution is expressed as
z = (x - u)/s
We want to determine the probability that no more than 25 were victims of e-mail fraud. It is expressed as
P(x lesser than or equal to 25)
The z value will be
z = (25- 35)/3.24 = - 10/3.24 = -3.09
Looking at the normal distribution table, the corresponding z score is 0.001
P(x lesser than or equal to 25) = 0.001
A trucking firm suspects that the mean lifetime of a certain tire it uses is more than 30,000 miles. To check the claim, the firm randomly selects and tests 54 of these tires and gets a mean lifetime of 29,400 miles with a population standard deviation of 1200 miles. At ΅ = 0.05, test the trucking firm's claim. Justify your decision with work. Write a short parargraph about the results of the test and what you can conclude about the claim
Answer:
We conclude that the lifetime of tires is less than 30,000 miles.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 30,000 miles
Sample mean, [tex]\bar{x}[/tex] = 29,400 miles
Sample size, n = 54
Alpha, α = 0.05
Population standard deviation, σ =1200 miles
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 30000\text{ miles}\\H_A: \mu < 30000\text{ miles}[/tex]
We use One-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{29400 - 30000}{\frac{1200}{\sqrt{54}} } = -3.6742[/tex]
Now, [tex]z_{critical} \text{ at 0.05 level of significance } = -1.64[/tex]
Since,
[tex]z_{stat} < z_{critical}[/tex]
We reject the null hypothesis and accept the alternate hypothesis.
Thus, we conclude that the lifetime of tires is less than 30,000 miles.
A pole that is 3.3 meters tall casts a shadow that is 1.69 meters long. At the same time, a nearby building casts a shadow that is 47.25 meters long. How tall is the building? Round to the nearest meter.
Answer:
92 meters
Step-by-step explanation:
This is a problem that can be solved by the concept of "similar triangles", where we have two right angle triangles that share also another common angle" the angle that the rays of the sun form with them (see attached image).
The smaller triangle is formed by the limiting rays of the sun, the pole, and its shadow: it has a height of 3.3 m (the length of the pole) and a base of 1.69 m (the pole's shadow).
The larger triangle is formed by the rays of the sun, the building and its shadow: it has a base of 47.25 m and an unknown height that we named as "x" (our unknown).
The ratio of the bases of such similar triangles must be in the same proportion as the ratio of their heights, so we can create a simple equation that equals such ratios, and then solve for the unknown "x":
[tex]\frac{H}{h} =\frac{B}{b}\\\frac{x}{3.3} =\frac{47.25}{1.69}\\x=\frac{47.25\,*\,3.3}{1.69} \\x=92.26331\, m[/tex]
which we can round to the nearest meter as: 92 m
Determine whether the lines L1:x=18+6t,y=7+3t,z=13+3t and L2:x=−14+7ty=−12+5tz=−8+6t intersect, are skew, or are parallel. If they intersect, determine the point of intersection; if not leave the remaining answer blanks empty.
Answer:
skew lines
Step-by-step explanation:
we are given 2 lines in parametric form as
L1:x=18+6t,y=7+3t,z=13+3t and L2:x=−14+7ty=−12+5tz=−8+6t[tex]L1:x=18+6t,y=7+3t,z=13+3t \\ L2:x=-14+7t,y=-12+5t,z=-8+6t[/tex]
If the lines intersect then the two points must be equal for one value of t.
Let us try equating x,y and z coordinate.
[tex]18+6t = -14+7t\\t=32\\[/tex]
when we equate y coordinate we get
[tex]7+3t =-12+5t\\2t =19\\t =9.5[/tex]
Since we get two different t we find that these two lines cannot intersect.
Comparing direction ratios we have
I line has direction ratios as (6,3,3) and second line (7,5,6)
These two are not proportional and hence not parallel
So these lines are skew lines
After analyzing the direction vectors of lines L1 and L2, it is clear that they are neither parallel nor do they intersect, which means the lines are skew.
Explanation:To determine whether the lines L1 and L2 intersect, are skew, or are parallel, we need to compare the direction vectors and check if they can be expressed as multiples of each other. If we consider the components of each direction vector given by the 't' terms in the equations of L1 and L2, they are (6, 3, 3) for L1 and (7, 5, 6) for L2 respectively.
To check for parallelism, we find a constant 'k' such that (6, 3, 3) = k*(7, 5, 6). After trying to solve for 'k', we immediately see that no such constant can exist, as 6/7 is not equal to 3/5 or 3/6. Therefore, these lines are not parallel.
To check for intersection, we need to find a common point that satisfies both line equations for some value of the parameter 't'. However, since an attempt to find such values results in inconsistent equations, it suggests that no such point exists and these lines do not intersect.
Given that the lines are neither parallel nor do they intersect, they must be skew lines. Skew lines are lines that do not intersect and are not parallel, lying in different planes.
The area under a particular normal curve between 6 and 8 is 0.695. A normally distributed variable has the same mean and standard deviation as the parameters for this normal curve. What percentage of all possible observations of the variable lie between 6 and 8?
Answer:
69.5%
Step-by-step explanation:
A feature of the normal distribution is that this is completely determined by its mean and standard deviation, therefore, if two normal curves have the same mean and standard deviation we can be sure that they are the same normal curve. Then, the probability of getting a value of the normally distributed variable between 6 and 8 is 0.695. In practice we can say that if we get a large sample of observations of the variable, then, the percentage of all possible observations of the variable that lie between 6 and 8 is 100(0.695)% = 69.5%.
In a poll of 1000 adults in July 2010, 540 of those polled said that schools should ban sugary snacks and soft drinks. Complete parts a and b below. a. Do a majority of adults (more than 50%) support a ban on sugary snacks and soft drinks? Perform a hypothesis test using a significance level of 0.05.State the null and alternative hypotheses. Note that p is defined as the population proportion of people who believe that schools should ban sugary foods.
Answer:
[tex]z=2.53[/tex]
[tex]p_v =P(z>2.53)=0.0057[/tex]
The p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so then we have enough evidence to reject the null hypothesis, and we can say that at 5% of significance, the proportion of adults who support a ban on sugary snacks and soft drinks is more than 0.5 or 50%.
Step-by-step explanation:
1) Data given and notation
n=1000 represent the random sample taken
X=540 represent the adults that said that schools should ban sugary snacks and soft drinks
[tex]\hat p=\frac{540}{1000}=0.54[/tex] estimated proportion of adults that said that schools should ban sugary snacks and soft drinks
[tex]p_o=0.5[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that majority of adults (more than 50%) support a ban on sugary snacks and soft drinks, the system of hypothesis are:
Null hypothesis:[tex]p\leq 0.5[/tex]
Alternative hypothesis:[tex]p > 0.5[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.54 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.53[/tex]
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a one side right tailed test the p value would be:
[tex]p_v =P(z>2.53)=0.0057[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so then we have enough evidence to reject the null hypothesis, and we can say that at 5% of significance, the proportion of adults who support a ban on sugary snacks and soft drinks is more than 0.5 or 50%.
At a large department store, the average number of years of employment for a cashier is 5.7 with a standard deviation of 1.8 years, and the distribution is approximately normal. If an employee is picked at random, what is the probability that the employee has worked at the store for over 10 years? 99.2% 0.8% 49.2% 1.7%
Answer:
option 0.8%
Step-by-step explanation:
Data provided in the question:
Mean = 5.7 years
Standard deviation, s = 1.8 years
Now,
P(the employee has worked at the store for over 10 years)
= P(X > 10 years)
= [tex]P (Z > \frac{X-Mean}{\sigma})[/tex]
or
= [tex]P (Z > \frac{10-5.7}{1.8})[/tex]
= P (Z > 2.389 )
or
= 0.008447 [from standard z table]
or
= 0.008447 × 100% = 0.84% ≈ 0.8%
Hence,
the correct answer is option 0.8%
To find the probability that an employee has worked at a large department store for over 10 years, we can use the z-score formula and a standard normal distribution table or calculator.
Explanation:To find the probability that an employee has worked at the store for over 10 years, we can use the z-score formula:
z = (x - μ) / σ
where x is the value we are interested in (10 years), μ is the mean (5.7 years), and σ is the standard deviation (1.8 years).
Plugging in the values, z = (10 - 5.7) / 1.8 = 2.39
Using a standard normal distribution table or a calculator, we can find that the probability of a z-score greater than 2.39 is approximately 0.008, or 0.8%.
Learn more about Probability here:https://brainly.com/question/32117953
#SPJ3