Ask question

Ask question

All categories

4 years ago

7

In a random sample of 28 families, the average weekly food expense was $95.60 with a standard deviation of $22.50. Determine whe

ther a normal distribution or a t-distribution should be used or whether neither of these can be used to construct a confidence interval. Assume the distribution of weekly food expenses is normally shaped.

1 answer:

likoan [24]4 years ago

3 0

Answer:

Step-by-step explanation:

Given that in a random sample of 28 families, the average weekly food expense was $95.60 with a standard deviation of $22.50.

Sample size =28<30

For a normal distribution to be used, the condition is sample size should be atleast 30 and the sample is strictly drawn at random.

Here we have sample size <30 hence t test can be used as the original distribution is normal.

To construct confidence interval t -distribution should be used

You might be interested in

Solve the given equation over the interval [0, 2π): cot^2 x − cot x.

AlekseyPX

$45^o, 90^o, 225^o, 270^o$
$\frac{ \pi }{4} , \frac{ \pi }{2} , \frac{5 \pi }{4} , \frac{3 \pi }{2}$

8 0

3 years ago

Q2.What is the ratio of volumes of the two cylinders formed by rolling a sheet of dimensions lxb in two different ways? Q3.What

Kobotan [32]

Answer:

2. b : l

3. 20cm

4. 49 $cm^{2}$

5. $(2\pi+1):2\pi$

Step-by-step explanation:

<u>Solution 2:</u>

Let cylinder is rolled along 'l':

Height of cylinder , h = length of rectangle = l

Perimeter of base = b

Let 'r' be the radius of cylinder's base:

$2\pi r = b\\\Rightarrow r = \dfrac{b}{2\pi}$

Volume of a cylinder is given as:

$V = \pi r^{2} h$

Putting the values:

$V_1 = \pi (\dfrac{b}{2\pi})^2 l\\\Rightarrow V_1 = (\dfrac{b^2}{4\pi}) l$

Let cylinder is rolled along 'b':

Height of cylinder , h = length of rectangle = b

Perimeter of base = l

Let 'r' be the radius of cylinder's base:

$2\pi r = l\\\Rightarrow r = \dfrac{l}{2\pi}$

Volume of a cylinder is given as:

$V = \pi r^{2} h$

Putting the values:

$V_2 = \pi (\dfrac{l}{2\pi})^2 b\\\Rightarrow V_2 = (\dfrac{l^2}{4\pi}) b$

Taking ratio:

$V_1:V_2 = \dfrac{(\dfrac{b^2}{4\pi}) l}{(\dfrac{l^2}{4\pi}) b} = b:l$

Solution 3:

Rectangle is rolled along its length to make a cylinder, so height will be equal to its length.

$\therefore$ height of cylinder = 20 cm

Solution 4:

Side of square = 7 cm

Height of cylinder =Side of square = 7 cm

7 cm will be the circumference of the circle.

i.e. $2\pi r$ = 7 cm

Curved surface area of a cylinder:

$CSA = 2\pi rh$

Putting the above values:

CSA = 7 $\times$ 7 = 49 $cm^{2}$

Solution 5:

As calculated in above step:

CSA = $2\pi rh =$ 7 $\times$ 7 = 49 $cm^{2}$

Total surface area = $2\pi r^{2} + 2\pi r h$

Calculating value of r:

$2\pi r$ = 7 cm

$\Rightarrow 2 \pi r = 7\\\Rightarrow r = \dfrac{7}{2\pi}$

Total surface area =

$2\pi (\dfrac{7}{2\pi})^{2} + 49\\\Rightarrow \dfrac{49}{2\pi}+49\\\Rightarrow 49(\dfrac{1}{2\pi}+1) cm^2$

Ratio of TSA: CSA is

$49(\dfrac{1}{2\pi}+1) cm^2 : 49 cm^2\\\Rightarrow (\dfrac{1}{2\pi}+1):1\\\Rightarrow (2\pi+1): 2\pi$

5 0

3 years ago

Lynn has 8 candy bars. She wants to give 1/4 of a candy bar to each person in her class. She has 30 kids in her class. Does she

nevsk [136]

Yes, Lynn does have enough for each classmate in class, If she wants to give each kid 1/4 of a candy bar, 4(the amount of kids that can get candy per bar) x8(candy bars) would equal 32.

5 0

3 years ago

Read 2 more answers

You invite your best friend Sheila out for a picnic on Thursday. Sheila tells you "If it rains, there is a 20% probability

den301095 [7]

Answer: 56%

Probably that it will rain and she goes + probably that it will be sunny and she goes

= 0.4 * 0.2 + 0.6*0.8

= 0.08 + 0.48

= 0.56

56% chance

3 0

3 years ago

Can someone help 2d + 1/5 = 3/5

bearhunter [10]

$2d+\dfrac{1}{5}=\dfrac{3}{5}\ \ \ \ \ |subtract\ \dfrac{1}{5}\ from\ both\ sides\\\\2d=\dfrac{2}{5}\ \ \ \ |divide\ both\ sides\ by\ 2\\\\\boxed{d=\frac{1}{5}}$

8 0

3 years ago