1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
kobusy [5.1K]
3 years ago
6

A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each on

e recorded. The results are given below. Assume the percentages of students' absences are approximately normally distributed. Use Excel to estimate the mean percentage of absences per tutorial over the past 5 years with 90% confidence. Round your answers to two decimal places and use increasing order.
Mathematics
1 answer:
SVEN [57.7K]3 years ago
7 0

Answer:

Step-by-step explanation:

Hello!

X: number of absences per tutorial per student over the past 5 years(percentage)

X≈N(μ;σ²)

You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.

The formula for the CI is:

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.

Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4

X[bar]= 10.41

S= 3.71

Z_{1-\alpha /2}= Z_{0.95}= 1.645

[10.41±1.645*(\frac{3.71}{\sqrt{28} } )]

[9.26; 11.56]

Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.

I hope this helps!

You might be interested in
2 y<br> 22 36<br> 27 54<br> 3272
Lelu [443]
What is the question itself?
4 0
3 years ago
At the Tucson Zoo, Jason spends $6.00 to enter the zoo and $2.00 per animal to feed them. Meanwhile, at the Phoenix Zoo, Kim spe
shtirl [24]

Answer:

3 animals

Step-by-step explanation:

  1. combine like terms
  • 6 + 2x = 15 + x
  • -6   +  x   - 6  - x

     --------------------------

         3x = 9

     2. divide both sides by 3

  • 3x/ 3 = 9/3
  • x = 3        
8 0
3 years ago
−10x+3y=5 ASAP<br> x=y−4 ​ <br><br> x ?<br> y?
nexus9112 [7]

Answer:

x = 1 and y = 5

Step-by-step explanation:

Use substitution because you know that x = y - 4, and plug this into the first equation to get -10(y - 4) + 3y = 5, or -10y + 40 + 3y = 5. This is -7y = -35 so y = 5. Plug this into the 2nd equation to get that x = 1 and y = 5.

3 0
2 years ago
A bag contains 20 marbles of the same size that are blue, green, and red. The probability of picking a blue marble is 20%. The p
kaheart [24]
The probability of picking up a red marble is 30%
3 0
3 years ago
Read 2 more answers
Approximating the equation of a line , can someone please help !
KatRina [158]

Answer:

y = -13x + 231

$94


3 0
3 years ago
Other questions:
  • A cabin has room for 7 campers and 2 conselors. How many cabins are needed for a total of 49 campers and 14 counselors?
    9·2 answers
  • You walk up to a tank of water that can hold up to 20 gallons. When it is active, a drain empties water from the tank at a const
    7·1 answer
  • The missing angle. or the one marked "?"
    8·1 answer
  • Y=19x+19b solve for b in the literal equation
    12·1 answer
  • WHAT capital letter forms a parallel lines
    11·2 answers
  • Suppose a new cancer treatment was given to a sample of 300 patients. The treatment was successful for 210 of the patients. Assu
    12·2 answers
  • How many distinct permutations of the letters of the word BANANA
    15·1 answer
  • Wich is a solution to (x-3)(x+9)=-27?​
    10·1 answer
  • Can anyone help me with this question . <br> Please do not answer if you do not know .
    8·2 answers
  • Which inequality is equivalent to 3+4/x&gt;_x+2/x
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!