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
oksian1 [2.3K]
3 years ago
9

A random sample of 18 graduates of a certain secretarial school typed an average of 80.6 words per minute with a standard deviat

ion of 7.2 words per minute. Assuming a normal distribution for the number of words typed per​ minute, compute the 95​% prediction interval for the next observed number of words per minute typed by a graduate of the secretarial school.
Mathematics
1 answer:
insens350 [35]3 years ago
7 0

Answer: ( 77.27 , 83.93)

Therefore at 95% confidence/prediction interval is

= ( 77.27 , 83.93)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = 80.6 words per minute

Standard deviation r = 7.2

Number of samples n = 18

Confidence interval = 95%

z(at 95% confidence) = 1.96

Substituting the values we have;

80.6+/-1.96(7.2/√18)

80.6+/-1.96(1.697056274847)

80.6 +/- 3.33

= ( 77.27 , 83.93)

Therefore at 95% confidence/prediction interval is

= ( 77.27 , 83.93)

You might be interested in
The tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm2 and a stand
Elanso [62]

Answer:

95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment.

Step-by-step explanation:

We are given that the tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm 2 and a standard deviation of 2.15.

A machine was recently adjusted and a sample of 50 items were taken to determine if the mean tensile strength has changed. The mean of this sample is 74.28. Assume that the standard deviation did not change because of the adjustment to the machine.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

                         P.Q. = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean strength of 50 items = 74.28

            \sigma = population standard deviation = 2.15

            n = sample of items = 50

            \mu = population mean tensile strength after machine was adjusted

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.</em>

So, 95% confidence interval for the population mean, \mu is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level of

                                                  significance are -1.96 & 1.96}  

P(-1.96 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.96) = 0.95

P( -1.96 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

P( \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

<u><em>95% confidence interval for</em></u> \mu = [ \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ]

                 = [ 74.28-1.96 \times {\frac{2.15}{\sqrt{50} } } , 74.28+1.96 \times {\frac{2.15}{\sqrt{50} } } ]

                 = [73.68 kg/mm2 , 74.88 kg/mm2]

Therefore, 95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

<em>Yes, this data suggest that the tensile strength was changed after the adjustment as earlier the mean tensile strength was 72 kg/mm2 and now the mean strength lies between 73.68 kg/mm2 and 74.88 kg/mm2 after adjustment.</em>

8 0
3 years ago
PLS HELP SP=3x+1, and LN=10x−6. Find SP.<br><br> A. 9<br> B. 4<br> C. 21<br> D. 7
Umnica [9.8K]

Step-by-step explanation:

2(sp)=ln

6x+2=10x-6

8=4x

x=2

therefore sp=3×2+1=6+1=7

Option D

5 0
3 years ago
You play a gambling game with a friend in which you roll a die. if a 1 or 2 comes up, you win $9. how much should you lose on an
EleoNora [17]
To make it a fair game you should lose $4.5
3 0
3 years ago
How do I solve this using elimination?<br><br> 80x + 45y = 645 <br> x + y = 12
sergeinik [125]
X + y = 12  / x (-45);
Then, -45x - 45y = -540;
We have 80x + 45y = 645 and
               -45x - 45y = -540
            ________________
                35x           = 105;
Finally, x = 105÷35;
x = 3;
y = 12 - 3;
y = 9;
7 0
3 years ago
Read 2 more answers
If a visitor only wants to taste 3 of the poultry items, and if the order of tasting is important, how many ways are there to do
lana [24]
This would be 3! ways

= 3*2*1  = 6 ways Answer
7 0
3 years ago
Other questions:
  • Determine whether the relation is a function: {(–1, –2), (–2, –3), (–3, –1)}
    6·1 answer
  • The number of students at parkville high school is 21 less than 2 times the number of students at midtown middel school which ex
    11·1 answer
  • Factor the polynomial using the pattern x^2-15x+54
    12·2 answers
  • Which equation could generate the curve in the graph below?
    15·1 answer
  • How would you know if a polynomial is prime?
    7·1 answer
  • Find the average of the following numbers. -100 and 102.
    15·2 answers
  • g An article reported the results of a survey of 400 elementary school teachers and 500 high school teachers. Of the elementary
    11·1 answer
  • The temperature in noon was 6 degrees. It dropped 1.2 degrees each hour for the next 12 hours. What was the temperature at midni
    8·1 answer
  • HELP PLEASE I WOULD REALLY APPRECIATE IT - Ken can run m miles
    9·1 answer
  • As used in lines 1-2, "not readily verifiable" most
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!