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Lisa [10]
3 years ago
7

Alan bowls one game, which consists of 10 frames. The number of pins he knocked down in each frame is 5, 6, 7, 5, 7, 9, 2, 4, 8,

and 7. Which measure should Alan use to learn the difference of the greatest and the least number of pins he knocked down?
Mathematics
2 answers:
tankabanditka [31]3 years ago
8 0

Answer:

Range.

Step-by-step explanation:

Alan must use the measure called Range, which indicates the dispersion between the extreme values of a variable. It is calculated as the difference between the highest and the lowest value of the variable. It is denoted as R. First the data is sorted, then it is calculated as:

R = x (n) - x (1) where,

x (n): It is the largest value of the variable.

x (1): It is the smallest value of the variable.

In the given case (5, 6, 7, 5, 7, 9, 2, 4, 8, and 7), we sort the data, and get,

2, 4, 5, 5, 6, 7, 7, 7, 8, 9

x (n) = 9

X (1) = 2

R = 9 - 2 = 7

Hope this helps!

mote1985 [20]3 years ago
7 0
Alan shoud use mean and range, mean is where you add all of the numbers up and divide by how many numbers there are, and range where you take the highest number and subtract it by the lowest number. This way he will find the greatest and least number of pins he knocked down 
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5 0
3 years ago
A quality control manager is concerned about variability of the net weight of his company’s individual yogurt cups. To check the
jok3333 [9.3K]

Answer:

a) We reject H₀

b) The manager won´t be satisfied with nominal filling its cup

c) See step-by-step explanation

Step-by-step explanation:

Normal distribution  n <  30, therefore, we should use t - student table

Sample size  n  =  16

degree of freedom  =   df =  n - 1     df = 15

Sample mean    μ  = 5,85 ou

Sample standard deviation  is    s = 0,2 ou

Hypothesis test

Null hypothesis                               H₀              μ   >=  μ₀

Alternative hypothesis                   Hₐ               μ   <   μ₀

CI  = 95 %    then   α = 5 %     α = 0,05     α/2  =  0,025

Then in t-student table we find  t(c) = 1,753

To calculate  t(s)

t(s) =  (   μ   -  μ₀  ) s/√n

t(s)  =   ( 5,85 -  6 ) / 0,2/√16

t(s)  =  -  0,15* 4 / 0,2

t(s)  =  -  3

To compare  t(s) and t(c)

|t(s)| > |t(c)|        3 > 1,753

Then t(s) is in the rejection region. We should reject H₀. Data indicate that at 95 % of CI μ seems to be smaller than 6 ou

b) The manager won´t be satisfied with nominal filling its cup

4 0
3 years ago
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