A survey of all medium and large-sized corporations showed that 64% of them offer retirement plans to their employees. Let P be the proportion in a random sample of 50 such corporations that offer retirement plans to their employees. Find the probabilty that the value of P will be: a. less than 0.57 b. greater than 0.71

Answer:

The probability is [tex]\frac{5}{9}[/tex]

Step-by-step explanation:

The total number of students are 200.

number of full timers is 110 and number of part timers is 90.

number of male students is 90 and number of female students is 110.

Let the probability of part timers be P(B).

P(B) = [tex]\frac{90}{200}[/tex] = [tex]\frac{9}{20}[/tex]

Let the probability of female part timers be P(A)

P(A) = [tex]\frac{50}{200} = \frac{5}{20}[/tex]

now, the final probability is

= [tex]\frac{P(A)}{P(B)}[/tex]

=[tex]\frac{5/20}{9/20} = \frac{5}{9}[/tex]

Mrs. Maxwell needs to buy 18.66 packs of pencil containing 336 pencils for 84 students.

Given that  

Mrs. Maxwell is buying pencils for her students.

She has 3 classes of 28 students each.

She wants to give each of her students 4 pencils.

She can purchase packs of 18 pencils for $2.52

Need to determine number of packs of pencils Mrs. Maxwell need to purchase.

Let’s first determine total number of students  

As there are 3 classes of 28 students each means

Number of students in 1 class = 28

=> Number of student in 3 classes = 28 x 3 = 84

Mrs. Maxwell wants to give each of her students 4 pencils.

So number of pencils required for 1 student = 4

=> number of pencils required for 84 student = 4 x 84 = 336

Given that number of pencils in one pack = 18  

So number of pack containing 336 pencils [tex]=\frac{336}{18}=18.66 \text { packs }[/tex]

So Mrs. Maxwell needs to buy 18.66 packs of pencil containing 336 pencils for 84 students.

Final answer:

Mrs. Maxwell needs to purchase 19 packs of pencils to provide 4 pencils to each of her 84 students, as pencil packs come in sets of 18 and she needs a total of 336 pencils.

Explanation:

The question requires us to calculate the total number of pencil packs Mrs. Maxwell needs to purchase for her students for Valentine’s Day. Mrs. Maxwell has 3 classes with 28 students each, which totals to 3 * 28 = 84 students. Since she wants to give each student 4 pencils, she will need 84 * 4 = 336 pencils in total. Pencil packs come in sets of 18, so to find out how many packs she will need, we divide the total number of pencils by the number in each pack: 336 pencils ÷ 18 pencils per pack = 18.67 packs. Since she cannot buy a fraction of a pack, she needs to round up to the nearest whole number, which is 19 packs of pencils.

Answer:

a) Mean = 6.5, sample standard deviation = 3.50

b) Standard error = 0.7826

c) Point estimate = 6.5

d) Confidence interval:  (5.1469 ,7.8531)

Step-by-step explanation:

We are given the following data set for students to earn bachelor's degrees.

4, 4, 4, 4, 4, 4, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 6, 6, 8, 9, 9, 13, 13, 15

a) Formula:

[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]  

where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]Mean =\displaystyle\frac{130}{20} = 6.5[/tex]

Sum of squares of differences = 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 4 + 4 + 4 + 4 + 4 + 4 + 0.25 + 0.25 + 2.25 + 6.25 + 6.25 + 42.25 + 42.25 + 72.25 = 233.5

[tex]S.D = \sqrt{\frac{233.5}{19}} = 3.50[/tex]

b) Standard Error

[tex]= \displaystyle\frac{s}{\sqrt{n}} = \frac{3.50}{\sqrt{20}} = 0.7826[/tex]

c) Point estimate for the mean time required for all college is given by the sample mean.

[tex]\bar{x} = 6.5[/tex]

d) 90% Confidence interval:  

[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]  

Putting the values, we get,  

[tex]t_{critical}\text{ at degree of freedom 19 and}~\alpha_{0.10} = \pm 1.729[/tex]  

[tex]6.5 \pm 1.729(\frac{3.50}{\sqrt{20}} ) = 6.5 \pm 1.3531 = (5.1469 ,7.8531)[/tex]

e) No, the confidence interval does not contain the value of 4 years. Thus, confidence interval is not a good estimator as most of the value in the sample is of 4 years. Most of the sample does not lie in the given confidence interval.

Final answer:

The question requires calculating the sample mean, standard deviation, standard error, point estimate for the population mean, and constructing a confidence interval for the mean time to earn bachelor's degrees. The value of 4 years will need to be checked against the calculated confidence interval, and the reliability of these results may be scrutinized based on the distribution of the data.

Explanation:

The question involves calculating various statistical measures for a dataset representing the number of years college students took to earn bachelor's degrees. To address these parts:

To calculate the sample mean, you add up all the numbers and divide by the total count of numbers. The sample standard deviation measures the amount of variation or dispersion in a set of values. You use the formula for standard deviation for a sample.

The standard error (SE) is calculated by dividing the sample standard deviation by the square root of the sample size.

The point estimate for the mean time is the sample mean, as it provides the best estimate of the population mean based on the sample data.

To construct the confidence interval, you would typically use the sample mean ± (critical value from the t-distribution * standard error). For 90% confidence, you can find the critical t-value for 19 degrees of freedom (since sample size minus one equals degrees of freedom).

To answer whether the confidence interval contains the value of 4 years and discuss the reliability of this interval, you'll examine the calculated interval and consider factors like the presence of outliers and the shape of the distribution.

Answer: 6

Step-by-step explanation:

Let  S= Total colleges

A = colleges are expensive.

B= colleges are  far from home( more than 200 miles away).

Given : n(S)= 20

n(A)=8

n(B)=8

n(A∩B) =2

Then, the number of  college that are not expensive and within 200 miles from home :-

[tex]n(A'\cap B')=n(S)-n(A\cup B)\\\\=20-(n(A)+n(B)-N(A\cap B))\ \ [\because\ n(A\cup B)=n(A)+n(B)-N(A\cap B)]\\\\=20-(8+8-2)\\\\=20-14=6[/tex]

i.e.  the number of  college that are not expensive and within 200 miles from home=6

Hence, the number of possible selections are 6 .

Melanie, as a researcher, needs to sample at least 139 full-term newborn babies at her hospital to be 95% confident that the mean birthweight of the sample is within 100 grams of the mean of all babies.

To estimate the mean birthweight of full-term babies in her hospital with an error of at most 100 grams and a 95% confidence level, Melanie can use the formula for sample size in a normal population: n = (Z^2 * σ^2) / E^2 where Z is the Z-value from the Z-table for the desired level of confidence (for 95%, Z = 1.96), σ is the standard deviation of the population (600 grams), and E is the maximum allowable error (100 grams).

Plugging in these values, we get n = (1.96^2 * 600^2) / 100^2 = 138.2976, which we round up to 139 since we can't have a fractional number of babies.

So, Melanie should sample at least 139 babies to be at least 95% confident that the mean birthweight of the sample is within 100 grams of the mean birthweight of all babies.

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Answer:

atleast 385

Step-by-step explanation:

Given that an electrical power company is looking to expand into a new market. Before they commit to supplying the new area with electricity, they would like know the mean daily power usage for homes there.

Population std deviation = [tex]\sigma = 50[/tex]

Sample size =[tex]n[/tex]

STd error of sample mean = [tex]\frac{50}{\sqrt{n} }[/tex]

Margin of error for 95% would be Critical value ( std error)

Here since population std dev is known we can use Z critical value= 1.96

[tex]1.96*\frac{50}{\sqrt{n} }<5\\n>19.6^2\\n>384.16[/tex]

Sample size should be atleast 385

The confidence interval for population mean (when population standard deviation is unknown) is given by :-

[tex]\overline{x}-t^*\dfrac{s}{\sqrt{n}}< \mu<\overline{x}+z^*\dfrac{s}{\sqrt{n}}[/tex]

, where n= sample size

[tex]\overline{x}[/tex] = Sample mean

s= sample size

t* = Critical value.

Given : n= 25

Degree of freedom : [tex]df=n-1=24[/tex]

[tex]\overline{x}= \$93.36[/tex]

[tex]s=\ $19.95[/tex]

Significance level for 98% confidence interval : [tex]\alpha=1-0.98=0.02[/tex]

Using t-distribution table ,

Two-tailed critical value for 98% confidence interval :

[tex]t^*=t_{\alpha/2,\ df}=t_{0.01,\ 24}=2.4922[/tex]

⇒ The critical value that should be used in constructing the confidence interval = 2.4922

Then, the 95% confidence interval would be :-

[tex]93.36-(2.4922)\dfrac{19.95}{\sqrt{25}}< \mu<93.36+(2.4922)\dfrac{19.95}{\sqrt{25}}[/tex]

[tex]=93.36-9.943878< \mu<93.36+9.943878[/tex]

[tex]=93.36-9.943878< \mu<93.36+9.943878[/tex]

[tex]=83.416122< \mu<103.303878\approx83.4161<\mu<103.3039[/tex]

Hence, the 98% confidence interval for the mean repair cost for the dryers. = [tex]83.4161<\mu<103.3039[/tex]

Answer:

Step-by-step explanation:

Answer:

Since the p–value is less than the significance level, the null hypothesis is rejected. The true proportion of engineering students planning graduate studies significantly different from 0.5 at  α=0.10

Step-by-step explanation:

Please see attachment

Answer: The open interval would be (31.4,42.5).

Step-by-step explanation:

Since we have given that

mean = 36.9

Standard deviation = 16.5

n = 48

At 98% confidence interval, z = 2.33

So, Interval would be

[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=36.9\pm 2.33\dfrac{16.5}{\sqrt{48}}\\\\=36.9\pm 5.549\\\\=(36.9-5.5,36.9+5.6\\\\=(31.4,42.5)[/tex]

Hence, the open interval would be (31.4,42.5).

To find the 98% confidence interval for the number of seeds, use the formula for confidence intervals and the values given. The 98% confidence interval for the number of seeds is (31.1, 42.7).

To find the 98% confidence interval for the number of seeds, we can use the formula:

Confidence Interval = mean ± (critical value) * (standard deviation / sqrt(sample size))

Since we want a 98% confidence interval, the critical value is found using the z-table. It is approximately 2.33.

Plugging in the values:

Confidence Interval = 36.9 ± (2.33) * (16.5 / sqrt(48))

Simplifying the expression gives us the 98% confidence interval for the number of seeds as (31.1, 42.7).

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We are using a t-test for hypothesis testing given a population mean, sample mean, sample size, and sample standard deviation. After calculations, the test-statistic was found to be approximately 2.88.

This is a question that involves statistical hypothesis testing. We're given a population mean (μ = 7.4), a sample mean (x = 8.6), the sample size (n = 31), and the sample standard deviation (s = 2.9). We are asked to test the claim that the new drug changes the blood's pH level using a 5% significance level.

This is a case of a two-tailed test, because we're interested in whether the drug changes, which means it could be either increase or decrease the blood's pH level.

To answer part (b) of the question, you would use a t-distribution as your sampling distribution. In many cases, especially in health sciences, Student's t-distribution is used when the sample size is less than 30 and the population standard deviation is not known. It is also used when the sample follow a normal distribution or when the sample size is large.

For the test statistic using the t-distribution, we would use the formula: t = (x - μ) / (s / sqrt(n)). Plugging in our numbers, we get: t = (8.6 - 7.4) / (2.9 / sqrt(31)), which yields a test-statistic of approximately 2.88 when rounded to three decimal places.

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This hypothesis test utilises t-distribution, given the small sample size. First, find the standard error (SE) via dividing the standard deviation by the root of the sample size. Then, calculate the t-statistic by dividing the means difference by the SE.

The question is related to the field of statistics. Specifically, it is about Hypothesis Testing. The hypothesis we are testing is whether the mean pH in blood has been changed by a new drug from its accepted norm of 7.4.

Because we have a small sample size (n<30), we will use the t-distribution for our hypothesis test. This is due to the Central Limit Theorem, which states that for samples of size 30 or more, the sampling distribution will approximate a normal distribution. But for less than 30, we should use the t-distribution.

To calculate the test statistic:

This is the t-value you use to test your hypothesis on a t-distribution curve.

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Answer:

249.6×432.6=107,976.96

Answer:

Step-by-step explanation:

2.4*104=249.6

4.2*103=432.6

(249.6)(432.6)=107976.96

All a matter of simple multiplication. ;)

Answer:

Step-by-step explanation:

Since 6 cakes serve 120 people, each cake serves 120/6 = 20 people.  Each cake weighs (5 lb)(16 oz/lb) = 80 oz. Then each person gets ...

  (80 oz)/(20 persons) = 4 oz/person

__

150 servings will require 150/20 = 7.5 cakes. Peter already has 6 cakes, so needs to make 2 more.

5lb per cake *6 cakes= 30lb

30lb / 120 people = 0.25lb per person

150 people - 120 people = 30 people

0.25lb per person * 30 people = 7.5lb

5lb___1 cake

7.5lb___X=1.5 cakes

(7.5lb * 1 cake)/5lb = 1.5 cakes

Final answer:

Calculating the probability of different card events when dealt a hand of three cards involves understanding basic principles of probability and operating with fractions. Each event requires calculating the probability of successive draws, considering that cards are dealt without replacement, impacting the probability of each subsequent draw.

Explanation:

Let's address each part of the question dealing with probabilities when being dealt a hand of three cards from a standard 52-card deck:

The expected number of children who get his or her own coat among 10 students using expected deviation will be 6.

Probability is defined as the possibility of the occurrence of an event.

Probability lies between 0 and 1.

A low standard deviation suggests that data are grouped around the mean, whereas a large standard deviation shows that data are more dispersed.

Given that:

Number of students = 10

The probability for 10 students is 0.1

The pay-off table is as follows:

The expected deviation can be calculated as:

[tex]E(x) = 1\times0.1 + 2\times0.1+ 3\times0.1+ 4\times0.1 + 5\times0.1 + 6\times0.1 + 7\times0.1 + 8\times0.1 + 9\times0.1 + 10\times0.1[/tex]

E(x) = 5.5

The expected number of children who get his or her own coat is 6 when rounded off.

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The table showing the relation between the number of students to its probability is shown below.

Final answer:

The expected number of children who will get their own coat when the coats are handed out randomly to 10 students is 6, based on the concept of expected value in probability.

Explanation:

To calculate the expected number of children who get their own coat among 10 students using expected deviation:

First, assign a probability to each possible outcome (number of children getting their own coat). Since there are 10 students and each has an equal probability of getting their own coat, the probability for each outcome is 0.1.

Next, multiply each outcome by its respective probability and sum them up. This gives the expected value (E(x)).

E(x) = (1 * 0.1) + (2 * 0.1) + (3 * 0.1) + (4 * 0.1) + (5 * 0.1) + (6 * 0.1) + (7 * 0.1) + (8 * 0.1) + (9 * 0.1) + (10 * 0.1)

E(x) = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 + 0.8 + 0.9 + 1.0

E(x) = 5.5

When rounded off, the expected number of children who get their own coat is 6.

Answer:

[693,38:725,62]hs

Step-by-step explanation:

Hello!

Your study variable is X: the lifespan of a light bulb. This variable is said to have an approximately normal distribution.

X≈N(μ;σ²)

Were

μ= populatiom mean

σ= 45 hs standard deviation

A sample of n= 35

To estimate the population mean with a confidence interval you have to use the Z statistic:

X{bar} ± [tex]Z_{1-\alpha /2}[/tex]*(σ/√n)

[tex]Z_{1-\alpha /2} = Z_{0.98} = 2.054[/tex]

710 ± 2.054*(45/√35

[693,38:725,62]hs

I hope you have a SUPER day!

Answer:  C

Step-by-step explanation:

Rejecting the null hypothesis means we've found a significant difference in the means.  That means the probability that we'd see means so far apart by chance is less than our threshold of significance.

The probability that the ant will be at 1 after 3 seconds is 49/64.

To find the probability that the ant will be at 1 after 3 seconds, we need to consider all possible paths it can take. After each second, the ant can either move left with a probability of 1/4 or move right with a probability of 3/4.

Let's analyze all the possible paths:

Adding up the probabilities from each path, the total probability that the ant will be at 1 after 3 seconds is (27/64) + (9/64) + (9/64) + (3/64) + (1/64) = 49/64.

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Answer:  95% confidence interval would be (0.449, 0.511)

Step-by-step explanation:

Since we have given that

n = 1022

p = 48% = 0.48

We need to find the 95% confidence interval first.

z = 1.96

Margin of error would be

[tex]z\sqrt{\dfrac{p(1-p)}{n}}\\\\=1.96\times \sqrt{\dfrac{0.48\times 0.52}{1022}}\\\\=0.031[/tex]

95% confidence interval would be

[tex]p\pm 0.031\\\\=(0.48-0.031,0.48+0.031)\\\\=(0.449,0.511)[/tex]

It means true proportion who felt that economic growth is more important than protecting the environment is within 0.449 and 0.511 using 95% confidence.

Answer:

a₁, a₂, a₃, . . .

Step-by-step explanation:

When we name something in Mathematics it is always advisable to number them as [tex]$ a_1, a_2, a_3, ...$[/tex].

Because the alphabets are only 26 in number we might run out of notations.

The simple logic behind the notion of [tex]$ a_1, a_2, a_3, ... $[/tex] is that the numbers have no end and we can number as many variables we want using this logic.

In this problem, you can continue your notation from [tex]$ a_{27}, a_{28}, a_{29} $[/tex] so that you don't have to make a change in your figure.

Answer: 500 miles

Step-by-step explanation:

Given : Sergio and Lizeth have planned to rent a car from a company that charges $75 a week plus $0.25 a mile.

i.e. Fixed charge= $75

Rate per mile = $0.25

Let x denotes the number of miles.

Then, Total charges = Fixed charge+ Rate per mile x No. of miles traveled

=  $75+ $0.25x

To keep budget within $200, we have following equation.

[tex]75+0.25x=200\\\\\Rightarrow\ 0.25=200-75\\\\\Rightarrow\ 0.25=125\\\\\Rightarrow\ x=\dfrac{125}{0.25}=\dfrac{12500}{25}=500[/tex]

Hence, they can travel 500 miles and still keep within their $200 budget.

Answer:

If only one of these statements is true, #17 robbed the bank.

Step-by-step explanation:

1st Scenario: #26 tells the truth

If #26 is telling the truth, that means #14 is guilty and, consequently, lying. However, that would also mean that #17 is telling the truth and then more than one statement would be true.

2nd Scenario: #17 tells the truth

If #17 is telling the truth, #17 is innocent, but then again either #26 or #14 are lying and two statements would be true.

3rd Scenario: #14 tells the truth

If #14 is telling the truth, #14 is innocent and, consequently, #26 is lying. That leaves us with #17 claiming innocence, but since only #14 can be telling the truth, #17 is lying and robbed the bank.

Answer:

Step-by-step explanation:

Hello!

The objective of this experiment is to test if a new feed + additive generates a better production of milk in cows. For this, the owner selects 13 cows and randomly separates them into two groups.

Group 1 has 8 cows that receive the new feed + additive.

Group 2 has 5 cows that were fed with the old feed.

After two weeks of feeding the animals with the different feeds, the production of milk of each group was recorded so that they can be compared.

Since you have two separate groups to wich at random two different treatments were applied and later the variable was measured, these two samples/groups are independent and the proper test to compare the population means of the milk production in both groups is a pooled t.

I hope this helps!

Answer:

Mean = [tex]\mu = 1140[/tex]

Standard deviation = [tex]\sigma = 80[/tex]

Find the probabilities for the average length of life of the selected batteries.

A)The average is between 1128 and 1140.

We are supposed to fidn P(1128<x<1140)

Formula : [tex]Z=\frac{x-\mu}{\sigma}[/tex]

At x = 1128

[tex]Z=\frac{1128-1140}{80}[/tex]

[tex]Z=-0.15[/tex]

Refer the z table for p value

P(x<1128)=0.4404

At x = 1140

[tex]Z=\frac{1140-1140}{80}[/tex]

[tex]Z=0[/tex]

Refer the z table for p value

P(x<1140)=0.5

So,P(1128<x<1140)=P(x<1140)-P(x<1128)=0.5-0.4404=0.0596

Hence the probabilities for the average length of life of the selected batteries  is between 1128 and 1140 is 0.0596

B)The average is greater than 1152.

P(x>1152)

At x = 1128

[tex]Z=\frac{1152-1140}{80}[/tex]

[tex]Z=0.15[/tex]

Refer the z table for p value

P(x<1152)=0.5596

So,P(x>1152)=1-P(x<1152)=1-0.5596=0.4404

Hence the probabilities for the average length of life of the selected batteries is greater than 1152 is 0.4404

C) The average is less than 940.

P(x<940)

At x = 940

[tex]Z=\frac{940-1140}{80}[/tex]

[tex]Z=-2.5[/tex]

Refer the z table for p value

P(x<940)=0.5596

Hence the probabilities for the average length of life of the selected batteries is less than 940 is 0.5596

The question involves calculating the probability of the lifespan of car batteries based on given statistical data using the standard normal distribution.

The student is asking about calculating probabilities related to the life expectancy of car batteries using the principles of statistics. They provided information about the average lifespan of the batteries (1140 days), the standard deviation (80 days), and the sample size (400).

For part (a):

To find the probability that the average lifespan of the selected batteries is between 1128 and 1140 days, you would use the sampling distribution of the sample mean. The mean of the sampling distribution is the same as the population mean, 1140, and the standard deviation (often termed the standard error) of this distribution is the population standard deviation divided by the square root of the sample size: σ/√n or 80/√400 = 4 days. You would then use a standard normal distribution to find the probability that a normally distributed random variable falls between the z-scores corresponding to 1128 and 1140 days.

For part (b):

To determine the probability that the average is greater than 1152, again find the z-score for 1152 and use the standard normal distribution.

For part (c):

Asking for the probability of the average being less than 940 is a theoretical scenario far from the mean. Given the standard deviation, this would likely yield a probability close to 0.

Answer:

a) [tex]\bar{x} = 18.21[/tex]

b) 1.64

c) (16.57,19.85)

d) The sample size must be 135 or greater if the error must not exceed $1.00.

Step-by-step explanation:

We are given the following information in the question:

Sample size, n = 50

Sample mean =   $18.21

population standard deviation = $5.92

a)  Point estimate for the population mean cost

[tex]\bar{x} = 18.21[/tex]

b) Margin of error =

[tex]z_{critical}\displaystyle\frac{\sigma}{\sqrt{n}}[/tex]

[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]

Margin of error =  [tex]1.96\displaystyle\frac{5.92}{\sqrt{50}} = 1.64[/tex]

c) 95% Confidence interval

[tex]\mu \pm z_{critical}\displaystyle\frac{\sigma}{\sqrt{n}}[/tex]

Putting the values, we get,

[tex]18.21 \pm 1.64) = (16.57,19.85)[/tex]

d) Marginal error less than $1.00

[tex]1.96\displaystyle\frac{\sigma}{\sqrt{n}} \leq 1\\\\\sqrt{n} \geq 1.96\times 5.92\\n \geq (11.6)^2\\n \geq 134.56 \approx 135[/tex]

Thus, the sample size must be 135 or greater if the error must not exceed $1.00.

(A) Point estimate for the population mean cost: $18.21. (b) Margin of error: $1.64. (c) 95% Confidence interval: $16.57 ≤ μ ≤ $19.85. (d) The sample size must be 135 or greater if the error must not exceed $1.00.

(a) Point estimate for the population mean cost

The point estimate for the population mean (μ) is the sample mean ([tex]\bar{x}[/tex]). In this case:

Point estimate ([tex]\bar{x}[/tex]) = $18.21

(b) Margin of error

The formula for the margin of error (E) in a confidence interval is given by:

E = Z * (σ / √n)

where:

Z is the Z-score corresponding to the desired confidence level,

σ is the population standard deviation,

n is the sample size.

For a 95% confidence interval, Z is approximately 1.96. Let's plug in the values:

E ≈ 1.96 * (5.92 / √50)

E ≈ 1.96 * (5.92 / 7.07)

E ≈ 1.64

So, the margin of error is approximately $1.64.

(c) 95% Confidence interval

The confidence interval is given by:

Confidence interval = [tex]\bar{x}[/tex] - E ≤ μ ≤ [tex]\bar{x}[/tex] + E

Plugging in the values:

$18.21 - 1.64 ≤ μ ≤ $18.21 + 1.64

$16.57 ≤ μ ≤ $19.85

So, the 95% confidence interval for the population mean cost is $16.57 ≤ μ ≤ $19.85.

(d) Sample size needed if the error must not exceed $1.00

The formula for the margin of error is:

E = Z * (σ / √n)

We want the error (E) to be less than $1.00, so:

$1.00 = 1.96 * (5.92 / √n)

Solving for n:

√n = 1.96 * (5.92 / 1.00)

n = [(1.96 * 5.92) / 1.00]^2

For a 95% confidence interval, Z is approximately 1.96:

n ≈ [(1.96 * 5.92) / 1.00]^2

n ≈ (11.5872 / 1.00)^2

n ≈ (11.5872)^2

n ≈ 134.52

Since the sample size must be a whole number, we round up to the nearest whole number. Therefore, n = 135.

So, the correct answer for (d) is:

The sample size must be 135 or greater if the error must not exceed $1.00.

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Answer: Margin of error would be 1.47 for 95% confidence.

Step-by-step explanation:

Since we have given that

Mean = 26.8

Standard deviation = 7.5

n = 100

We need to find the margin of error for 95% confidence.

So, z = 1.96

So, the margin of error would be

[tex]z\times \dfrac{\sigma}{\sqrt{n}}\\\\=1.96\times \dfrac{7.5}{\sqrt{100}}\\\\=\dfrac{14.7}{10}\\\\=1.47[/tex]

Hence, margin of error would be 1.47 for 95% confidence.

The margin of error for a 95% confidence interval for the mean BMI of 100 young women sampled from the NHANES data with a standard deviation of 7.5 would be approximately ±1.47.

The margin of error for a 95% confidence interval is determined using the standard deviation and the sample size. Given that you have a standard deviation (σ) of 7.5 and a sample size (n) of 100, you can calculate the standard error (SE) using the formula SE = σ/√n. To compute your 95% margin of error, multiply the standard error by the z-score associated with a 95% confidence level, which is 1.96.

Apply these to the formulas:

So, the margin of error for a 95% confidence interval for the mean BMI of 100 young women sampled would be approximately ±1.47.

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Final answer:

The function y = 1.29x + 3 represents the price y a website store charges for shipping x items. The reasonable range for this function would be all positive real numbers.

Explanation:

The function y = 1.29x + 3 represents the price y a website store charges for shipping x items. The reasonable range for this function depends on the context. Since the function represents the price of shipping, the range should be positive, as shipping cannot have a negative price.

Therefore, the reasonable range for this function would be J- all positive real numbers.

Final answer:

The reasonable range for the shipping cost function starts from 3 and includes values that increase by 1.29 for each additional item, corresponding to whole numbers of items shipped. Therefore, the answer is G - {4.29, 5.58, 6.87, ...}.

Explanation:

The function given is y = 1.29x + 3, which represents the price y that a website store charges for shipping x items. Since shipping cannot have a negative cost and the minimum number of items shipped is either zero or a positive integer, the reasonable range for this function would begin at the point where x is zero. Therefore, we start our range by calculating the shipping cost for zero items: 1.29(0) + 3 = 3. As x increases, the cost will also increase linearly according to the function. Hence, all subsequent shipping prices will be greater than 3.

Now, let's consider what the reasonable range for shipping items would be. It would be abnormal to have a fractional number of items shipped because items are discrete entities. Thus, the list of shipping prices should only include charges for whole numbers of items. So, the range should only include prices that correspond to whole numbers of items shipped.

As a result, the reasonable range would include values starting from y = 3 onwards at intervals of 1.29 times an integer value. Option G, which starts from 4.29 and increases at a constant rate of 1.29, represents these intervals since 4.29 is the price for shipping one item (1.29*1 + 3). Any positive number of items shipped will result in a corresponding shipping price that is greater than 3, and since it is discreetly incremented, the prices will form a sequence of specific numbers, not all positive real numbers. Therefore, option G - {4.29, 5.58, 6.87, ...} - is the most appropriate answer.

Answer:

y=3 or y-3=0

Step-by-step explanation:

-2y=-6

-2y÷(-2)=-6÷(-2)

y=-6÷(-2)

y=-6÷2

y=3

or

-2y=-6

-2y+6=0

-y+3=0

y-3=0

See picture for answer and solution steps.

Answer:  The required position function is [tex]s(t)=t^3+t^2-5t+7.[/tex]

Step-by-step explanation:  Given that a particle moves in a straight line and has acceleration given by

[tex]a(t)=6t+2.[/tex]

The initial velocity of the particle is v(0) = −5 cm/s and its initial displacement is s(0) = 7 cm.

We are to find the position function s(t).

We know that the acceleration function a(t) is the derivative of the velocity function v(t). So,

[tex]v^\prime(t)=a(t)\\\\\Rightarrow v^\prime(t)=6t+2\\\\\Rightarrow v(t)=\int (6t+2) dt\\\\ \Rightarrow v(t)=3t^2+2t+A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

Also, the velocity function v(t) is the derivative of the position function s(t). So,

[tex]s^\prime(t)=v(t)\\\\\Rightarrow s^\prime(t)=3t^2+2t+A\\\\\Rightarrow s(t)=\int(3t^2+2t+A) dt \\\\\Rightarrow s(t)=t^3+t^2+At+B~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]

From equation (i), we get

[tex]v(0)=0+0+A\\\\\Rightarrow A=-5,~\textup{where A is a constant}[/tex]

and from equation (ii), we get

[tex]s(0)=0+0+0+B\\\\\Rightarrow B=7,~\textup{where B is a constant}.[/tex]

Substituting the values of A and B in equation (ii), we get

[tex]s(t)=t^3+t^2-5t+7.[/tex]

Thus, the required position function is [tex]s(t)=t^3+t^2-5t+7.[/tex]

Velocity: [tex]\( v(t) = 3t^2 + 2t - 5 \)[/tex] cm/s. Position: [tex]\( s(t) = t^3 + t^2 - 5t + 7 \)[/tex] cm, starting at 7 cm with initial velocity of -5 cm/s.

let's solve this step by step.

Given that ( a(t) = 6t + 2 ), we need to find the velocity function ( v(t) ) by integrating the acceleration function with respect to time.

1. **Find velocity function ( v(t) )**:

[tex]\[ a(t) = \frac{dv}{dt} \][/tex]

  So, integrating ( a(t) ) with respect to ( t ) will give us ( v(t) ):

[tex]\[ \int a(t) \, dt = \int (6t + 2) \, dt \][/tex]

[tex]\[ v(t) = \int (6t + 2) \, dt = 3t^2 + 2t + C_1 \][/tex]

  Here, ( C_1 ) is the constant of integration.

  Given that ( v(0) = -5 ) cm/s, we can find the value of ( C_1 ):

[tex]\[ v(0) = 3(0)^2 + 2(0) + C_1 = C_1 = -5 \][/tex]

  So, [tex]\( v(t) = 3t^2 + 2t - 5 \).[/tex]

2. **Find position function ( s(t) )**:

  We know that velocity is the rate of change of displacement. So, we need to integrate the velocity function with respect to time to find the position function.

[tex]\[ v(t) = \frac{ds}{dt} \][/tex]

  Integrating ( v(t) ) with respect to ( t ) will give us ( s(t) ):

[tex]\[ \int v(t) \, dt = \int (3t^2 + 2t - 5) \, dt \][/tex]

[tex]\[ s(t) = \int (3t^2 + 2t - 5) \, dt = t^3 + t^2 - 5t + C_2 \][/tex]

  Here, [tex]\( C_2 \)[/tex] is the constant of integration.

  Given that [tex]\( s(0) = 7 \)[/tex] cm, we can find the value of [tex]\( C_2 \):[/tex]

[tex]\[ s(0) = (0)^3 + (0)^2 - 5(0) + C_2 = C_2 = 7 \][/tex]

  So, [tex]\( s(t) = t^3 + t^2 - 5t + 7 \).[/tex]

Therefore, the position function of the particle is [tex]\( s(t) = t^3 + t^2 - 5t + 7 \)[/tex]cm.

Answer:

0.3446,0.2119

Step-by-step explanation:

Given that Jill’s bowling scores are approximately normally distributed with mean 170 and standard deviation 20, while Jack’s scores are approximately normally distributed with mean 160 and standard deviation 15.

If X represents Jill scores and Y Jack scores we have

X is N(170,20) and Y is N(160,15)

since x and y are independent we have the difference

X-Y is [tex]N(170-16,\sqrt{20^2+15^2} )\\=N(10, 25)[/tex]

a) Prob that Jack’s score is higher

= P(-x+y>0)

=[tex]P(Z<\frac{10}{25} )\\= P(Z<0.4)\\\\= =0.3446[/tex]

b) X+Y is Normal with (330, 25)

[tex]P(X+Y>350) = P(Z>\frac{350-330}{25} )\\=P(Z>0.8)\\\\= =0.2119[/tex]

To find the probability that Jack's score is higher than Jill's, calculate the z-scores and compare them. For the probability that the total of their scores is above 350, find the z-score for the sum and use a standard normal distribution table. The probability of Jack's score being higher is 50% and the probability of the sum being above 350 is 78.81%.

To approximate the probability that Jack's score is higher than Jill's, we can use the concept of the z-score. The z-score measures how many standard deviations a value is from the mean. For Jack's score, we calculate his z-score as (his score - his mean) / his standard deviation, which is (160 - 160) / 15 = 0. For Jill's score, the z-score is (170 - 170) / 20 = 0.

Since both z-scores are 0, we can conclude that the probability of Jack's score being higher than Jill's is 0.5, or 50%.

To find the probability that the total of their scores is above 350, we need to find the z-score for the sum. The sum of their scores is 160 + 170 = 330. The mean of the sum is 160 + 170 = 330, and the standard deviation of the sum is sqrt((15^2) + (20^2)) = sqrt(625) = 25.

Therefore, the z-score for the sum is (350 - 330) / 25 = 0.8. Using a standard normal distribution table or calculator, we can find that the probability of the sum being above 350 is approximately 0.7881, or 78.81%.

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