Answer:
The typical error between a predicted college GPA using this regression model and an actual college GPA for a given student will be about 0.374 grade points in size (absolute value).
Step-by-step explanation:
The linear regression line for College GPA based on High school GPA is:
College GPA = 1.07 + 0.62 High-school GPA
It is provided that the standard error of the regression line is,
[tex]s_{e}=0.374[/tex]
The standard error of a regression line is the average distance between the predicted value and the regression equation.
It is the square root of the average squared deviations.
It is also known as the standard error of estimate.
The standard error of 0.374 implies that:
The typical error between a predicted college GPA using this regression model and an actual college GPA for a given student will be about 0.374 grade points in size (absolute value).
The standard error of the regression tells us the average amount that the actual college GPA deviates from the predicted college GPA. A smaller standard error indicates a better fit of the model to the data. In this case, the small standard error suggests that the linear regression model provides a good prediction of college GPA based on high school GPA for the students at the local college.
Explanation:The standard error of the regression tells us the average amount that the actual college GPA deviates from the predicted college GPA based on the high school GPA. In this case, the standard error of the regression is 0.374. This means that, on average, the actual college GPA for a student deviates from the predicted college GPA by approximately 0.374.
This value gives us an idea of the accuracy of the linear regression model in predicting college GPA based on high school GPA. A smaller standard error indicates a better fit of the model to the data, implying that the predicted college GPA is closer to the actual college GPA. Conversely, a larger standard error suggests that the model's predictions are less accurate.
In this case, the standard error of the regression is relatively small (0.374), which indicates that the linear regression model provides a good prediction of college GPA based on high school GPA for the students at the local college.
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Consider the parameterization of the unit circle given by x=cos(3t^2-t), y=sin(3t^2-t) for t in (-infinity, infinity). Describe in words and sketch how the circle is traced out, and use this to answer the following questions.
(a) When is the parameterization tracing the circle out in a clockwise direction? _________?
(Give your answer as a comma-separated list of intervals, for example, (0,1), (3,Inf)). Put the word None if there are no such intervals.
(b) When is the parameterization tracing the circle out in a counter-clockwise direction? ______?
(Give your answer as a comma-separated list of intervals, for example, (0,1), (3,Inf)). Put the word None if there are no such intervals.
(c) Does the entire unit circle get traced by this parameterization?
A. yes
B. no
(d) Give a time t at which the point being traced out on the circle is at (10):
t= ___________?
Answer and Step-by-step explanation:
The answer is attached below
In this exercise we have to use the knowledge of parameterization and calculate the direction and direction of the equation, so we have to:
A) Clockwise: [tex]t \in [ -\infty, 1/6][/tex]
B) Counter-clockwise: [tex]t \in [ 1/6, \infty][/tex]
C) [tex]\theta \in [ 0, 2 \pi][/tex]
D) [tex]t= 0 \ or \ t=1/3[/tex]
For this exercise, the following equations were informed:
[tex]x= cos(3t^2-t)\\y= sin(3t^2-t)\\t \in [ -\infty, \infty][/tex]
taking the parameterization we have that:
[tex]\phi = 3t^2 - t= t(3t-1)[/tex]
As t increases from [tex][ -\infty, \infty][/tex] [tex]\phi[/tex] decreases, after 0 it becomes negative and after 1/3, goes on increasing. Also:
[tex]\frac{d\phi}{dt} = (6t-1)\\t= 1/6[/tex]
a) For clockwise begin [tex]\phi[/tex] must be decreasing, so:
[tex]t \in [ -\infty, 1/6][/tex]
b) For counter-clockwise [tex]\phi[/tex] must be increasing, so:
[tex]t \in [ 1/6, \infty][/tex]
c) Entise circle gets traced out. For we know:
[tex]x= cos\theta\\y= sin\theta[/tex]
Circle gets traced out once for:
[tex]\theta \in [ 0, 2 \pi][/tex]
d) When point (1, 0) so:
[tex]1= cos(3t^2-t)\\0= sin(3t^2-t)\\t= 0 \or \ t=1/3[/tex]
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A few weeks into the deadly SARS (Severe Acute Respiratory Syndrome) epidemic in 2003, the number of cases was increasing by about 4% each day.† On April 1, 2003, there were 1,804 cases. Find an exponential model that predicts the number of cases t days after April 1, 2003. f(t) = Use it to estimate the number of cases on April 26, 2003. (The actual reported number of cases was 4,836.)
Answer:
[tex]f(t)=1804(1.04)^{t}\\f(25)=4809.17[/tex]
Step-by-step explanation:
1. Since the increasing rate is 0.04 or (4%) per day, then the factor is (1+0.04) raised to t days, and we have and exponential growth therefore we can write:
[tex]\\f(t)=c(1.04)^t\\f(t)=1,804(1.04)^t\\[/tex]
2. To estimate the number of cases, 25 days later following that exponential model
[tex]f(25)=1804(1.04)^{25}\\f(25)=4809.17[/tex]
Can some help me with this question
Answer:
Step-by-step explanation:
A quadrilateral inscribed within a circle is known as a cyclic quadrilateral. Property of the cyclic quadrilateral is that, sum of its opposite angles is 180°.
∴ ∠U + ∠K = 180°
∴ ∠K = 180 - 85 = 95°
The average heights of a random sample of 400 people from a city is 1.75 m. It is known that the heights of the population are random variables that follow a normal distribution with a variance of 0.16.
Determine the interval of 95% confidence for the average heights of the population.
Answer:
The 95% confidence interval for the average heights of the population is between 1.7108m and 1.7892m.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]
Now, find M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
The standard deviation is the square root of the variance. So
[tex]\sigma = \sqrt{0.16} = 0.4[/tex]
Then
[tex]M = 1.96*\frac{0.4}{\sqrt{400}} = 0.0392[/tex]
The lower end of the interval is the mean subtracted by M. So it is 1.75 - 0.0392 = 1.7108m
The upper end of the interval is the mean added to M. So it is 1.75 + 0.0392 = 1.7892m
The 95% confidence interval for the average heights of the population is between 1.7108m and 1.7892m.
Answer:
The interval of 95% confidence for the average heights of the population is = [tex](1.7108, 1.7892)[/tex]Step-by-step explanation:
mean x = [tex]1.75[/tex]
Variance [tex]\rho^2 = 0.16[/tex]
standard deviation [tex](\rho) = \sqrt{0.16} = 0.4[/tex]
n = 400
[tex]95\%[/tex] confidence :
[tex]\alpha = 100\% - 95\% = 5\%\\\\\frac{\alpha}{2} = 2.5\% = 0.025[/tex]
From standard normal distribution table,
[tex]Z_\frac{\alpha}{2} = Z_{0.025} = 1.96[/tex]
Margin of error, [tex]E = Z_\frac{\alpha}{2} * \frac{\rho}{\sqrt{n}}[/tex]
[tex]E = 1.96 * \frac{0.4}{\sqrt{400}}\\\\E = 0.0392[/tex]
Lower limit: x - E
[tex]= 1.75 - 0.0392\\\\= 1.7108[/tex]
Upper limit: x + E
[tex]= 1.75 + 0.0392\\\\= 1.7892[/tex]
[tex]Limits : (1.7108, 1.7892)[/tex]
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If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Write function that take a number n and return the sum of all the multiples of 3 or 5 below n.
Answer:
The function is provided below.
Step-by-step explanation:
Running the program in Python, the code for this problem is as follows:
def func(n):
count = 0 **initializing the sum with 0**
for i in range (1, n):
if i%3 == 0 or i%5 == 0:
count = count + 1
return (count)
When the program is executed enter value 1 to 10 one by one and the result will be 23, the sum of all the multiples of 3 and 5.
Match the scenario with its distribution type. Group of answer choices Sammy takes a sample of 200 students at her college in an attempt to estimate the average time students spend studying at her university. She plots her data using a histogram, the histogram shows what type of distribution? Phil wants to estimate the average driving time to work for individuals in Corvallis, Oregon. He takes a simple random sample of 100 individuals and records their drive time. Phil knows that a different sample will yield a different sample average. Suppose Phil was able to repeatedly sample a different group of Corvallis residents until he obtained every combination of 100 residents. If he plots the sample means from each sample this distribution will represent which type of distribution? An instructor wants see the scores of her entire class after the midterm. The distribution of scores represents what type of distribution?
Answer:
Scenario 1: Sample distributionScenario 2: Sampling distribtuionScenario 3: Population distributionStep-by-step explanation:
Scenarios are: Sample Distribution, sampling distribution, Population distribution
sample distribution: A sample distribution contains the data of individuals of sample collected from a population.
Sampling distribution: distribution of multiple sample statistics
Population distribution: Statistics of entire population without drawing sny sample
In scenario 1, only one saple is taken from a population and that sample is plotted
In Scenario 2, multiple samples are taken from a population and means of each sample is plotted
In Scenario 3, scores of entire population is considered.
9. A large electronic office product contains 2000 electronic components. Assume that the probability that each component operates without failure during the useful life of the product is 0.995, and assume that the components fail independently. Approximate the probability that five or more of the original 2000 components fail during the useful life of the product.
Answer:
97.10% probability that five or more of the original 2000 components fail during the useful life of the product.
Step-by-step explanation:
For each component, there are only two possible outcomes. Either it works correctly, or it does not. The probability of a component falling is independent from other components. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
In this problem we have that:
[tex]n = 2000, p = 1-0.995 = 0.005[/tex]
Approximate the probability that five or more of the original 2000 components fail during the useful life of the product.
We know that either less than five compoenents fail, or at least five do. The sum of the probabilities of these events is decimal 1. So
[tex]P(X < 5) + P(X \geq 5) = 1[/tex]
We want [tex]P(X \geq 5)[/tex]
So
[tex]P(X \geq 5) = 1 - P(X < 5)[/tex]
In which
[tex]P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{2000,0}.(0.005)^{0}.(0.995)^{2000} = 0.000044[/tex]
[tex]P(X = 1) = C_{2000,1}.(0.005)^{1}.(0.995)^{1999} = 0.000445[/tex]
[tex]P(X = 2) = C_{2000,2}.(0.005)^{2}.(0.995)^{1998} = 0.002235[/tex]
[tex]P(X = 3) = C_{2000,3}.(0.005)^{3}.(0.995)^{1997} = 0.007480[/tex]
[tex]P(X = 4) = C_{2000,4}.(0.005)^{4}.(0.995)^{1996} = 0.018765[/tex]
[tex]P(X < 5) = P(X = 0) + `P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.000044 + 0.000445 + 0.002235 + 0.007480 + 0.018765 = 0.0290[/tex]
[tex]P(X \geq 5) = 1 - P(X < 5) = 1 - 0.0290 = 0.9710[/tex]
97.10% probability that five or more of the original 2000 components fail during the useful life of the product.
A certain paper suggested that a normal distribution with mean 3,500 grams and a standard deviation of 560 grams is a reasonable model for birth weights of babies born in Canada.
One common medical definition of a large baby is any baby that weighs more than 4,000 grams at birth.
What is the probability that a randomly selected Canadian baby is a large baby?
The probability that a randomly selected Canadian baby is a large baby (weighing more than 4,000 grams) is approximately 0.187 or 18.7%.
Explanation:To find the probability that a randomly selected Canadian baby is a large baby, we need to calculate the area under the normal distribution curve to the right of 4,000 grams. First, we calculate the z-score using the formula: z = (x - mean) / standard deviation. Plugging in the values, we get z = (4000 - 3500) / 560 = 0.8929.
Next, we need to find the area under the curve to the right of this z-score using a standard normal distribution table or a calculator. The cumulative probability from the table or calculator is approximately 0.187. This means that the probability of a randomly selected Canadian baby being a large baby (weighing more than 4,000 grams) is approximately 0.187 or 18.7%.
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The weight of adobe bricks for construction is normally distributed with a mean of 3 pounds and a standard deviation of 0.25 pound. Assume that the weights of the bricks are independent and that a random sample of 25 bricks is chosen.
a) What is the probability that the mean weight of the sample is less than 3.10 pounds? Round your answer to four decimal places
b) What value will the mean weight exceed with probability 0.99? Round your answer to two decimal places.
Answer:
a) 0.9772 = 97.72% probability that the mean weight of the sample is less than 3.10 pounds
b) 2.88 pounds
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 3, \sigma = 0.25, n = 25, s = \frac{0.25}{\sqrt{25}} = 0.05[/tex]
a) What is the probability that the mean weight of the sample is less than 3.10 pounds? Round your answer to four decimal places
This is the pvalue of Z when X = 3.10. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{3.1 - 3}{0.05}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772
0.9772 = 97.72% probability that the mean weight of the sample is less than 3.10 pounds
b) What value will the mean weight exceed with probability 0.99? Round your answer to two decimal places.
This is the value of X when Z has a pvalue of 1-0.99 = 0.01. So X when Z = -2.33.
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]-2.33 = \frac{X - 3}{0.05}[/tex]
[tex]X - 3 = -2.33*0.05[/tex]
[tex]X = 2.88[/tex]
Solve the equation M=7r2h/19 for r in terms of M and h. Assume r, M and h are all positive.
Answer:
[tex]r=\frac{19M}{14h}[/tex]
Step-by-step explanation:
The equation is given as:
[tex]M=\frac{7r2h}{19}[/tex]
Assuming all the unknown variables are positive, we can make [tex]r[/tex] the subject of the formula to obtain it in terms of M & h:
[tex]M=\frac{7r2h}{19}\\M\times19=7r2h\\\\\frac{19M}{2h}=7r\\\\r=\frac{19M}{2h\times7}\\\\r=\frac{19M}{14h}[/tex]
or [tex]r=1.3571M/h[/tex]
Hence, r as in terms of M& H is given as
[tex]r=\frac{19M}{14h} \ or \ 1.3571M/h[/tex]
A manufactured lot of buggy whips has 20 items, of which 5 are defective. A random sample of 5 items is chosen to be inspected. Find the probability that the sample contains exactly one defective item
Answer:
[tex] P(X=1)[/tex]
And using the probability mass function we got:
[tex] P(X=1) = (5C1) (0.25)^1 (1-0.25)^{5-1}= 0.396[/tex]
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
For this cae that one buggy whip would be defective is [tex] p = \frac{5}{20}=0.25[/tex]
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=5, p=0.25)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
And we want to find this probability:
[tex] P(X=1)[/tex]
And using the probability mass function we got:
[tex] P(X=1) = (5C1) (0.25)^1 (1-0.25)^{5-1}= 0.396[/tex]
A sampling distribution refers to the distribution of:
A. a sample
B. a population
C. a sample statistic
D. a population parameter
E. repeated samples
F. repeated populations
Answer:
The answer is a population parameter.
Step-by-step explanation:
Population can include people, but other examples include objects, event, businesses, and so on. Population is the entire pool from which statistical sample is drawn.
A parameter is a value that describes a characteristics of an entire population, such as population mean, because you can almost never measure an entire population, you usually don't know the real value of a parameter.
Consider all possible sample of size N that can be drawn from a given population (either with or without replacement). For example, we can compute a statistics (such as the mean and the standard deviation ) that will vary from sample to sample. In this manner we obtain a distribution of statistics that is called Sampling distribution.
In statistics, a sampling distribution is the theoretical distribution of a sample statistic that arises from drawing all possible samples of a specific size from a population. It helps to quantify the variability and predictability of sample statistics when used as estimates for population parameters.
Explanation:A sampling distribution refers to the "distribution of a sample statistic". This is option C from your list. This term describes the probability distribution of a statistic based on a random sample. For example, if we study random samples of a certain size from any population, the mean score will form a distribution. This is the sampling distribution of the mean. Similarly, variance, standard deviations and other statistics also have sampling distributions. The purpose of a sampling distribution is to quantify the variation and uncertainty that arises when we use sample statistics (like the mean) to estimate population parameters (like the population mean).
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Let P(n) be the statement that a postage of n cents can be formed using just 4-cent and 7-cent stamps. Use strong induction to prove that P(n) is true for all integers greater than or equal to some threshold x.
Answer:
True for n = 18, 19, 20, 21
Step-by-step explanation:
[tex]P(n) =[/tex] a postage of [tex]n[/tex] cents; where [tex]P(n) = 4x + 7y[/tex]. ( [tex]x[/tex] are the number of 4-cent stamps and [tex]y[/tex] are the number of 7-cent stamps)
For [tex]n=18, P(18)[/tex] is true.
This is a possibility, if [tex]x= 1 \ and \ y=2[/tex]
[tex]P(18) = 4(1) + 7(2) = 4 + 14 = 18[/tex]
Similarly for [tex]P(19)[/tex]:
[tex]P(19) = 4(3) + 7(0) = 19[/tex]
[tex]P(20) = 4(5) + 7(0) = 20\\P(21) = 4(0) + 7(3) = 21[/tex]
Consider the probability that no less than 92 out of 159 students will pass their college placement exams. Assume the probability that a given student will pass their college placement exam is 55%. Approximate the probability using the normal distribution. Round your answer to four decimal places.
Answer:
0.2843 = 28.43% probability that no less than 92 out of 159 students will pass their college placement exams.
Step-by-step explanation:
I am going to use the binomial approximation to the normal to solve this question.
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:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]p = 0.55, n = 159[/tex]. So
[tex]\mu = E(X) = 159*0.55 = 87.45[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{159*0.55*0.45} = 6.27[/tex]
Probability that no less than 92 out of 159 students will pass their college placement exams.
No less than 92 is more than 91, which is 1 subtracted by the pvalue of Z when X = 91. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{91 - 87.45}{6.27}[/tex]
[tex]Z = 0.57[/tex]
[tex]Z = 0.57[/tex] has a pvalue of 0.7157
1 - 0.7157 = 0.2843
0.2843 = 28.43% probability that no less than 92 out of 159 students will pass their college placement exams.
To approximate the probability that at least 92 out of 159 students will pass their college placement exams using the normal distribution, we will follow these steps:
Step 1: Determine the parameters of the binomial distribution.
In a binomial distribution the parameters are the number of trials (n) and the probability of success in a single trial (p). For this particular case:
n = 159 (the number of students)
p = 0.55 (the probability that a given student will pass)
Step 2: Calculate the mean (μ) and standard deviation (σ) of the binomial distribution.
The mean (μ) of a binomial distribution is given by:
μ = n * p
The standard deviation (σ) is given by:
σ = sqrt(n * p * (1 - p))
For our case:
μ = 159 * 0.55 ≈ 87.45
σ = sqrt(159 * 0.55 * (1 - 0.55)) ≈ sqrt(159 * 0.55 * 0.45) ≈ sqrt(71.775) ≈ 8.472 (rounded to three decimal places for intermediate calculation)
Step 3: Apply the continuity correction.
Because we're approximating a discrete distribution with a continuous one, we use a continuity correction. To find the probability of at least 92 students passing, we'll look for the probability of X > 91.5 (since X is a discrete random variable, X ≥ 92 is equivalent to X > 91.5 when using a continuous approximation).
Step 4: Calculate the z-score for the corrected value.
The z-score is calculated by taking the value of interest, applying the continuity correction, subtracting the mean, and then dividing by the standard deviation:
z = (X - μ) / σ
Using the corrected value:
z = (91.5 - 87.45) / 8.472 ≈ 0.4774 (rounded to four decimal places for the calculation)
Step 5: Find the corresponding probability.
The z-score tells us how many standard deviations away from the mean our value of interest is. To find the probability that at least 92 students pass (i.e., P(X ≥ 92), we need to find 1 - P(Z < z) because the normal distribution table or a calculator gives us P(Z < z), which is the probability of being less than a certain z value.
We are looking for the probability that z is greater than our calculated value, which is the upper tail of the distribution.
Step 6: Consult the standard normal distribution table or use a calculator.
The z-score of approximately 0.4774 corresponds to a percentage in the standard normal distribution. Since we want P(Z > z), we need to subtract P(Z < z) from 1. If we look up the probability for z=0.4774 in standard normal distribution tables or use a calculator, we find that P(Z < z) is approximately 0.6832.
Therefore, P(Z > z) = 1 - P(Z < z) = 1 - 0.6832 = 0.3168.
Step 7: Round the answer.
Rounding P(Z > z) = 0.3168 to four decimal places, the answer remains 0.3168.
So, the approximate probability that at least 92 out of 159 students will pass the exam, using the normal distribution approximation, is 0.3168 when rounded to four decimal places.
In previous years, the average number of sheets recycled per bin was 59.3 sheets, but they believe this number may have increase with the greater awareness of recycling around campus. They count through 79 randomly selected bins from the many recycle paper bins that are emptied every month and find that the average number of sheets of paper in the bins is 62.4 sheets. They also find that the standard deviation of their sample is 9.86 sheets.
What is the value of the test-statistic for this scenario? Round your answer to 3 decimal places.
What are the degrees of freedom for this t-test?
Answer:
There is enough evidence to support the claim that the average number of sheets recycled per bin was more than 59.3 sheets.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 59.3
Sample mean, [tex]\bar{x}[/tex] = 62.4
Sample size, n = 79
Alpha, α = 0.05
Sample standard deviation, s = 9.86
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 59.3\text{ sheets}\\H_A: \mu > 59.3\text{ sheets}[/tex]
We use one-tailed t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{62.4 - 59.3}{\frac{9.86}{\sqrt{79}} } = 2.7945[/tex]
Degree of freedom = n - 1 = 78
Now, [tex]t_{critical} \text{ at 0.05 level of significance, 78 degree of freedom } = 1.6646[/tex]
Since,
[tex]t_{stat} > t_{critical}[/tex]
We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.
Conclusion:
Thus, there is enough evidence to support the claim that the average number of sheets recycled per bin was more than 59.3 sheets.
g Each year the density of 7 species of Odonata (dragonflies and damselflies) is monitored in a wetland preserve. If the density of each species is to be compared with the density of every other species, how many comparisons must be made
Answer:
There are 21 comparisons to be made.
Step-by-step explanation:
The number of species of Odonata monitored every year is, n = 7.
It is provided that the density of each species is compared with each other.
The number of ways to compare the species (N) without repetition is:
[tex]N=\frac{n(n-1)}{2}\\=\frac{7(7-1)}{2}\\=\frac{7\times6}{2}\\=21[/tex]
Thus, there are 21 comparisons.
The 21 comparisons are as follows:
Specie 1 is compared with the remaining 6.
Specie 2 has already with he 1st so it is compared with the remaining 5.
Specie 3 has already with he 1st and 2nd so it is compared with the remaining 4.
Specie 4 has already with he 1st, 2nd and 3rd so it is compared with the remaining 3.
Specie 5 has already with he 1st, 2nd, 3rd and 4th so it is compared with the remaining 2.
Specie 6 has already with he 1st, 2nd, 3rd, 4th and 5th so it is compared with the remaining 1.
And the specie 7 has already been compared with the others.
Total number of comparisons = 6 + 5 + 4 + 3 + 2 + 1 = 21.
Records show that the average number of phone calls received per day is 9.2. Find the probability of between 2 and 4 phone calls received (endpoints included) in a given day.
Answer:
4.76% probability of between 2 and 4 phone calls received (endpoints included) in a given day.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given interval.
Records show that the average number of phone calls received per day is 9.2.
This means that [tex]\mu = 9.2[/tex].
Find the probability of between 2 and 4 phone calls received (endpoints included) in a given day.
[tex](2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4)[/tex]
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 2) = \frac{e^{-9.2}*(9.2)^{2}}{(2)!} = 0.0043[/tex]
[tex]P(X = 3) = \frac{e^{-9.2}*(9.2)^{3}}{(3)!} = 0.0131[/tex]
[tex]P(X = 4) = \frac{e^{-9.2}*(9.2)^{4}}{(4)!} = 0.0302[/tex]
[tex](2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4) = 0.0043 + 0.0131 + 0.0302 = 0.0476[/tex]
4.76% probability of between 2 and 4 phone calls received (endpoints included) in a given day.
Past records indicate that the probability of online retail orders
that turn out to be fraudulent is 0.08. Suppose that, on a given
day, 20 online retail orders are placed. Assume that the number of
online retail orders that turn out to be fraudulent is distributed as a
binomial random variable.
a. What are the mean and standard deviation of the number of online
retail orders that turn out to be fraudulent?
b. What is the probability that zero online retail orders will turn
out to be fraudulent?
c. What is the probability that one online retail order will turn out
to be fraudulent?
d. What is the probability that two or more online retail orders
will turn out to be fraudulent?
Answer:
a) Mean = 1.6, standard deviation = 1.21
b) 18.87% probability that zero online retail orders will turn out to be fraudulent.
c) 32.82% probability that one online retail order will turn out to be fraudulent.
d) 48.31% probability that two or more online retail orders will turn out to be fraudulent.
Step-by-step explanation:
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
The mean of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
In this problem, we have that:
[tex]p = 0.08, n = 20[/tex]
a. What are the mean and standard deviation of the number of online retail orders that turn out to be fraudulent?
Mean
[tex]E(X) = np = 20*0.08 = 1.6[/tex]
Standard deviation
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{20*0.08*0.92} = 1.21[/tex]
b. What is the probability that zero online retail orders will turn out to be fraudulent?
This is P(X = 0).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{20,0}.(0.08)^{0}.(0.92)^{20} = 0.1887[/tex]
18.87% probability that zero online retail orders will turn out to be fraudulent.
c. What is the probability that one online retail order will turn out to be fraudulent?
This is P(X = 1).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{20,1}.(0.08)^{1}.(0.92)^{19} = 0.3282[/tex]
32.82% probability that one online retail order will turn out to be fraudulent.
d. What is the probability that two or more online retail orders will turn out to be fraudulent?
Either one or less is fraudulent, or two or more are. The sum of the probabilities of these events is decimal 1. So
[tex]P(X \leq 1) + P(X \geq 2) = 1[/tex]
We want [tex]P(X \geq 2)[/tex]
So
[tex]P(X \geq 2) = 1 - P(X \leq 1)[/tex]
In which
[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]
From itens b and c
[tex]P(X \leq 1) = 0.1887 + 0.3282 = 0.5169[/tex]
[tex]P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.5169 = 0.4831[/tex]
48.31% probability that two or more online retail orders will turn out to be fraudulent.
The probability is an illustration of a binomial distribution.
The mean and the standard deviationThe given parameters are:
n = 20
p = 0.08
The mean is calculated as:
[tex]\bar x = np[/tex]
So, we have:
[tex]\bar x = 20 * 0.08[/tex]
[tex]\bar x = 1.6[/tex]
The standard deviation is calculated as:
[tex]\sigma = \sqrt{\bar x * (1 - p)[/tex]
This gives
[tex]\sigma = \sqrt{1.6 * (1 - 0.08)[/tex]
[tex]\sigma = 1.21[/tex]
Hence, the mean is 1.6 and the standard deviation is 1.21
The probability that zero online retail orders will turn out to be fraudulentThis is calculated as:
[tex]P(x) = ^nC_x * p^x * (1 - p)^{n-x}[/tex]
So, we have:
[tex]P(0) = ^{20}C_0 * 0.08^0 * (1 - 0.08)^{20 - 0}[/tex]
[tex]P(0) =0.1887[/tex]
The probability that zero online retail orders will turn out to be fraudulent is 0.1887
The probability that one online retail order will turn out to be fraudulentThis is calculated as:
[tex]P(x) = ^nC_x * p^x * (1 - p)^{n-x}[/tex]
So, we have:
[tex]P(1) = ^{20}C_1 * 0.08^1 * (1 - 0.08)^{20 - 1}[/tex]
[tex]P(1) =0.3281[/tex]
The probability that one online retail orders will turn out to be fraudulent is 0.3281
The probability that two or more online retail orders will turn out to be fraudulentThis is calculated as:
[tex]P(x\ge 2) = 1 - P(0) - P(1)[/tex]
So, we have:
[tex]P(x\ge 2) = 1 - 0.1887 - 0.3281[/tex]
[tex]P(x\ge 2) = 0.4832[/tex]
The probability that two or more online retail orders will turn out to be fraudulent is 0.4832
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A classic counting problem is to determine the number of different ways that the letters of "possession" can be arranged. Find that number. The number of different ways that the letters of "possession" can be arranged is nothing.
Number of possible ways to arrange letters of word "possession" = 75600.
Step-by-step explanation:
A classic counting problem is to determine the number of different ways that the letters of "possession" can be arranged. We have, word "POSSESSION" which have 10 letters! as P,O,S,S,E,S,S,I,O,N . We have to find the number of ways letters of word "possession" can be arranged , in order to do that we will use simple logic as:
At first, we have a count of 10 letters and 10 places vacant to fill letters.
For first place we have a choice of 10 letters , after putting some letter from all 10 letters in first place now we are left with 9 places and 9 letters having 9 choices , similarly we'll be having 8 places , 8 letters and 8 choices and so on...... Therefore, Number of possible ways to arrange letters of word "possession" = [tex]10.9.8.7.6.5.4.3.2.1[/tex] = [tex]10![/tex] ( 10 factorial ) , but there are 4 letters "s" are repeated and 2 letters "o" are repeated so we need to eliminate the similar combinations ∴ Number of possible ways = [tex]\frac{10!}{4!(2!)}[/tex] = 75600.
∴Number of possible ways to arrange letters of word "possession" = 75600
Suppose that your statistics professor tells you that the scores on a midterm exam were approximately normally distributed with a mean of 79 and a standard deviation of 6. The top 15% of all scores have been designated As. Your score is 89. Did you earn an A
Answer:
You earned an A.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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.
In this problem, we have that:
[tex]\mu = 79, \sigma = 6[/tex]
The top 15% of all scores have been designated As.
This means that if Z for the score has a pvalue of 1-0.15 = 0.85 or higher, the score is designated as A.
Your score is 89. Did you earn an A?
We have to find the pvalue of Z when X = 89. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{89 - 79}{6}[/tex]
[tex]Z = 1.67[/tex]
[tex]Z = 1.67[/tex] has a pvalue of 0.9525. So yes, you earned an A.
Solve for q. √3q + 2 = √5
Follow below steps:
To solve for q in the equation √3q + 2 = √5, we first isolate the term with q by subtracting 2 from both sides of the equation.
√3q = √5 - 2
Then we square both sides of the equation to remove the square root:
(√3q)² = ( √5 - 2 )²
3q = ( √5 - 2 )²
Now, expand the right side of the equation:
3q = 5 - 2√5 * 2 + 2²
3q = 5 - 4√5 + 4
3q = 9 - 4√5
Then, divide both sides of the equation by 3 to solve for q:
q = (9 - 4√5) / 3
So, the value of q is (9 - 4√5) / 3.
The value of q is [tex]\frac{9 - 4\sqrt{5}}{3}[/tex].
To solve for q, we'll isolate it by performing operations to both sides of the equation to get q by itself.
Given the equation:
[tex]\[ \sqrt{3q} + 2 = \sqrt{5} \][/tex]
Subtract 2 from both sides:
[tex]\[ \sqrt{3q} = \sqrt{5} - 2 \][/tex]
Now, to isolate q, we need to square both sides of the equation:
[tex]\[ (\sqrt{3q})^2 = (\sqrt{5} - 2)^2 \]\[ 3q = (\sqrt{5} - 2)^2 \]\[ 3q = 5 - 4\sqrt{5} + 4 \]\[ 3q = 9 - 4\sqrt{5} \][/tex]
Now, divide both sides by 3 to solve for q:
[tex]\[ q = \frac{9 - 4\sqrt{5}}{3} \][/tex]
So, [tex]\( q = \frac{9 - 4\sqrt{5}}{3} \).[/tex]
3/4(ad).. solve.. a=12 d=9
Answer:
hope it helps you see the attachment for further information
Answer:
81
Step-by-step explanation:
According to government data, 74% of employed women have never been married. Rounding to 4 decimal places, if 15 employed women are randomly selected: a. What is the probability that exactly 2 of them have never been married
Answer:
0.000001 = 0.0001% probability that exactly 2 of them have never been married
Step-by-step explanation:
For each employed women, there are only two possible outcomes. Either they have already been married, or they have not. The women are chosen at random, which means that the probability of a woman having been already married is independent from other women. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
74% of employed women have never been married.
This means that [tex]p = 0.74[/tex]
15 employed women are randomly selected
This means that [tex]n = 15[/tex]
a. What is the probability that exactly 2 of them have never been married
This is P(X = 2).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{15,2}.(0.74)^{2}.(0.26)^{13} = 0.000001[/tex]
0.000001 = 0.0001% probability that exactly 2 of them have never been married
Consider the following homogeneous differential equation. y dx = 2(x + y) dy Use the substitution x = vy to write the given differential equation in terms of only y and v.
Answer:
[tex]ydv = (v +2)dy\\[/tex]
Step-by-step explanation:
We are given the following differential equation:
[tex]y dx = 2(x + y) dy[/tex]
We have to substitute
[tex]x = vy[/tex]
Differentiating we get,
[tex]\dfrac{dx}{dy} = v + y\dfrac{dv}{dy}[/tex]
Putting value in differential equation, we get,
[tex]y dx = 2(x + y) dy\\\\y\dfrac{dx}{dy}=2(x+y)\\\\y(v+y\dfrac{dv}{dy}) = 2(vy + y)\\\\vy + y^2\dfrac{dv}{dy} = 2vy +2y\\\\y^2\dfrac{dv}{dy}=vy +2y\\\\y^2dv = y(v+2)dy\\ydv = (v +2)dy\\[/tex]
is the differential equation after substitution.
The given homogeneous differential equation y dx = 2(x + y) dy can be rewritten in terms of y and v using the substitution x = vy. The result is the differential equation y dv/dy = v.
Explanation:The given differential equation is y dx = 2(x + y) dy. To write his equation in terms of y and v using the substitution x = vy, we must first differentiate both sides of x = vy with respect to x to get 1 = v + y dv/dx. We rearrange this to get dx/dy = 1 / (v + y dv/dy). The original equation can now be rewritten after substituting these values, you will get y / (v + y dv/dy) = 2(v + y), simplifying, we get v = 2v + 2y, and after rearranging, we get y dv/dy = v. This is the differential equation in terms of v and y.
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The null hypothesis in ANOVA is that all means of all groups are the same. The alternative is that at least one pair of means is different. We compute an F-statistic to explore sources of variability in our data to conduct the omnibus ANOVA. Question: what do you expect to happen when the null hypothesis is true?
A. More between group variability
B. Less between group variability
Answer:
Correct option: B. Less between group variability
Step-by-step explanation:
The Analysis of Variance (ANOVA) test is performed to determine whether there is a significant difference between the different group mean.
The hypothesis is defined as:
H₀: There is no difference between the group means, i.e. μ₁ = μ₂ = ... = μₙ
Hₐ: At least one of the mean is different from the others, i.e. μ[tex]_{i}[/tex] ≠ 0.
The test statistic is defined as:
[tex]F=\frac{SS_{between}}{SS_{within}}[/tex]
If the null hypothesis is true then the test statistic will be small and if it is false then the test statistic will be large.
In this case it is provided that the null hypothesis is true.
This implies that:
[tex]SS_{between}<SS_{within}[/tex]
Implying that the sum of squares for between group variability is less than within group variability.
Thus, if the null hypothesis is true there will be less between group variability.
Suppose Albers Elementary School has 44 teachers and Bothel Elementary School has 74 teachers. If the total number of teachers at Albers and Bothel combined is 87, how many teachers teach at both schools?
Answer: 31
Step-by-step explanation:
This is solved Using the theory of sets.
Teachers in Albers school = 44
Teachers in Bothel school = 74
Let the number of teachers that work in both schools be denoted as "x"
This implies that:
Number of teachers in Albers only = 44 - x
Number of teachers in Bothel only = 74 - x
And total number of teachers in both schools = 87,
then
x + (44-x) + (74-x) = 87
118 -x = 87
31 = x
This means the number of teachers that teach in both schools = 31.
Answer: 31 teachers teach at both schools.
Let:
A = Albers Elementary School
B = Bothel Elementary School
According to the question:
n(A) = 44, n(B) = 74, n(A U B) = 87.
Using the union formula we get:
[tex]n(A \cup B) = n(A)+n(B)-n(A \cap B)\\87 = 44+74-n(A \cap B)\\n(A \cap B)=44+74-87\\n(A \cap B)=31[/tex]
So, 31 teachers teach at both schools.
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The rate at which a professional tennis player used carbohydrates during a strenuous workout was found to be 1.7 grams per minute. If a line were graphed showing time (in minutes) on the horizontal axis and carbohydrates used (in grams) on the vertical axis, what would be the slope of the line?
How many carbohydrates (in grams) would the athlete use in 40 minutes?
Answer:
m=1.7
C=68 gr
Step-by-step explanation:
Function Modeling
We are given a relationship between the carbohydrates used by a professional tennis player during a strenuous workout and the time in minutes as 1.7 grams per minute. Being C the carbohydrates in grams and t the time in minutes, the model is
[tex]C=1.7t[/tex]
The slope m of the line is the coefficient of the independent variable, thus m=1.7
The graph of C vs t is shown in the image below.
To find how many carbohydrates the athlete would use in t=40 min, we plug in the value into the equation
[tex]C=1.7\cdot 40=68\ gr[/tex]
The slope of the line representing the rate of carbohydrate usage is 1.7. Multiply this rate (1.7 grams per minute) by the time (40 minutes) to find the total carbohydrates used, which is 68 grams.
Explanation:The rate at which the tennis player uses carbohydrates is 1.7 grams per minute. In the context of a graph, this rate would represent the slope of the line. So, the slope of the line would be 1.7. Slope, in mathematics, is defined as the change in the y-value (vertical axis) divided by the change in the x-value (horizontal axis). Here, the rate of carbohydrate usage (1.7 grams per minute) is the change in the y-value (carbohydrates used) per change in x-value (time).
Now, you also want to know how many carbohydrates the athlete would use in 40 minutes. We know that the rate of carbohydrate usage is 1.7 grams per minute. So, to find the total amount of carbohydrates used in 40 minutes, you'd simply multiply the rate by the time:
1.7 grams/minute * 40 minutes = 68 grams
So, the athlete would use 68 grams of carbohydrates in 40 minutes.
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It is often said that your chances of winning the lottery if you buy a ticket are just slightly higher than if you don't buy one! Suppose a Lotto game consists of picking 6 of 48 numbers.
What is the probability of winning with the very first Lotto ticket you purchase?
Answer:
1/48 % or 6/48 % chance
Step-by-step explanation:
The probability of winning with the very first Lotto ticket you purchase of the Lotto game consisting of picking 6 of 48 numbers is 1/12271512 or approximately 0.0000000815.
What is a permutation?A permutation is a process of calculating the number of ways to choose a set from a larger set in a particular order.
If we want to choose a set of r items from a set of n items in a particular order, we find the permutation nPr = n!/(n-r)!.
What is a combination?A combination is a process of calculating the number of ways to choose a set from a larger set in no particular order.
If we want to choose a set of r items from a set of n items in no particular order, we find the combination nCr = n!/{(r!)(n-r)!}.
How do we solve the given question?In the question, we are asked to determine the probability of winning a lottery by picking 6 numbers from 48 numbers with the first ticket we purchase.
First, we need to calculate the number of combinations of choosing 6 numbers from 48 numbers. As we need to consider no particular order, we will use combinations,
48C6 = 48!/{(6!)(48-6)!} = 48!/(6!*42!) = (43*44*45*46*47*48)/(1*2*3*4*5*6*) (As 48! = 42!*43*44*45*46*47*48, and 42! cancels itself from the numerator and the denominator).
or, 48C6 = 12,271,512.
So, we get the number of combinations = 12,271,512.
We know that we will choose only one particular set of 6 numbers.
∴ The probability of winning on the very first ticket = 1/12,271,512 ≈ 0.0000000815
∴ The probability of winning with the very first Lotto ticket you purchase of the Lotto game consisting of picking 6 of 48 numbers is 1/12271512 or approximately 0.0000000815.
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A tortoise and a hare are competing in a 1600-meter race. The arrogant hare decides to let the tortoise have a 510-meter head start. When the start gun is fired the hare begins running at a constant speed of 9 meters per second and the tortoise begins crawling at a constant speed of 5 meters per second.
a. Define a function f to represent the tortoise's distance from the finish line (in meters) in terms of the number of seconds t since the start of the race.
b. Solve f(t)=0for t.
c. Define a function g to represent the hare's distance from the finish line (in meters) in terms of the number of seconds t since the start of the race.
d. Solve g(t)=0 for t
e. Who won the race?
Answer:
(a)f(t)=1090-5t
(b)f(0)=1090metres
(c)g(t)=1600-9t
(d)g(0)=1600metres
(e)The Hare
Step-by-step explanation:
Total Distance =1600 metres
(a)The tortoise has a 510m headstart and a speed of 5m/s
Distance=Speed X Time
Distance at 5m/s = 5t
Total Distance covered by the tortoise at any time t= 510+5t
Therefore, The tortoise's distance from the finish line (in meters) in terms of the number of seconds t since the start of the race is given as:
f(t)=1600-(510+5t)
f(t)=1090-5t
(b)f(0)=1090-(5X0)
=1090metres
(c)The hare has a speed of 9m/s
Distance=Speed X Time
Distance at 9m/s = 9t
Total Distance covered by the hare at any time t= 9t
Therefore, The hare's distance from the finish line (in meters) in terms of the number of seconds t since the start of the race is given as:
g(t)=1600-9t
(d)
g(0)=1600-(9X0)=1600metres
(e)The race is finished when the distance from the finish line=0
For the Tortoise
f(t)=1090-5t=0
1090=5t
t=218seconds
For the Hare
g(t)=1600-9t=0
9t=1600
t=177.8seconds
The hare takes a shorter time to reach the finish line so he won the race.
The average daily high temperature in June in LA is 77 degree F with a standard deviation of 5 degree F. Suppose that the temperatures in June closely follow a normal distribution. What is the probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June? How cold are the coldest 10% of the days during June in LA?
Answer:
11.51% probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June
The coldest 10% of the days during June in LA have high temperatures of 70.6F or lower.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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.
In this problem, we have that:
[tex]\mu = 77, \sigma = 5[/tex]
What is the probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June?
This probability is 1 subtracted by the pvalue of Z when X = 83. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{83 - 77}{5}[/tex]
[tex]Z = 1.2[/tex]
[tex]Z = 1.2[/tex] has a pvalue of 0.8849.
1 - 0.8849 = 0.1151
11.51% probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June
How cold are the coldest 10% of the days during June in LA?
High temperatures of X or lower, in which X is found when Z has a pvalue of 0.1, so whn Z = -1.28
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.28 = \frac{X - 77}{5}[/tex]
[tex]X - 77 = -1.28*5[/tex]
[tex]X = 70.6[/tex]
The coldest 10% of the days during June in LA have high temperatures of 70.6F or lower.
A) The probability of observing a temperature ≥ 83°F in LA during a randomly chosen day in June is;
p(observing a temperature ≥ 83°F) = 11.507%
B) The coldest 10% of the days during June in LA have temperatures;
Less than or equal to 70.592 °F
This question involves z-distribution which is given by the formula;
z = (x' - μ)/σ
We are given;
Average daily temperature; μ = 77 °F
Standard deviation; σ = 5 °F
Since the temperatures follow a normal distribution, then if we want to find the probability of observing a temperature ≥ 83°F, then;
x' = 83 °F
Thus;
z = (83 - 77)/5
z = 6/5
z = 1.2
Thus;
from online z-score calculator, p-value = 0.11507
Thus, p(observing a temperature ≥ 83°F) = 11.507%
B) We want to find out how cold the coldest 10% of the days during June in LA;
Thus, it means that p = 10% = 0.1
z-score at p = 0.1 from z-score tables is;
z = -1.28155
Thus;
-1.28155 = (x' - 77)/5
-1.28155*5 = x' - 77
-6.40775 = x' - 77
x' = 77 - 6.40775
x' ≈ 70.592 °F
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