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PilotLPTM [1.2K]
3 years ago
10

On a quiz worth 5 points, four students earned a 5, seven students earned a 4, six students earned a 3, two students earned a 2,

and one student earned a zero. Find the average score. (round to the tenths place)
Mathematics
1 answer:
Keith_Richards [23]3 years ago
5 0

Answer:

The average score is 3.5.

Step-by-step explanation:

The average of a data set is the middle value or the most typical value of the set. It computed by dividing the sum of observations by the number of observations.

The sample space of the scores are:

S = {5, 5, 5, 5, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 2, 2, 0}

Total number of students = 4 +7 + 6 + 2 + 1 = 20.

The average score is:

Average=\frac{1}{n}\sum x\\=\frac{1}{20}(5+5+5+5+4+4+4+4+4+4+4+3+3+3+3+3+3+2+2+0)\\=\frac{70}{20} \\=3.5

Thus, the average score is 3.5.

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Step-by-step explanation:

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3 0
2 years ago
The tensile strength of a metal part is normally distributed with mean 40 pounds and standard deviation 5 pounds. If 50,000 part
nalin [4]

Answer:

a) how many would you expect to fail to meet a minimum specification limit of 35-pounds tensile strength?

7933 parts

b) How many would have a tensile strength in excess of 48 pounds?

2739.95 parts

Step-by-step explanation:

The formula for calculating a z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

a) how many would you expect to fail to meet a minimum specification limit of 35-pounds tensile strength?

z = (x-μ)/σ

x = 35 μ = 40 , σ = 5

z = 35 - 40/5

= -5/-5

= -1

Determining the Probability value from Z-Table:

P(x<35) = 0.15866

Converting to percentage = 15.866%

We are asked how many will fail to meet this specification

We have 50,000 parts

Hence,

15.866% of 50,000 parts will fail to meet the specification

= 15.866% of 50,000

= 7933 parts

Therefore, 7933 parts will fail to meet the specifications.

b) How many would have a tensile strength in excess of 48 pounds?

z = (x-μ)/σ

x = 48 μ = 40 , σ = 5

z = 48 - 40/5

z = 8/5

z = 1.6

P-value from Z-Table:

P(x<48) = 0.9452

P(x>48) = 1 - P(x<48)

1 - 0.9452

= 0.054799

Converting to percentage

= 5.4799%

Therefore, 5.4799% will have an excess of (or will be greater than) 48 pounds

We are asked, how many would have a tensile strength in excess of 48 pounds?

This would be 5.4799% of 50,000 parts

= 5.4799% × 50,000

= 2739.95

Therefore, 2739.95 parts will have a tensile strength excess of 48 pounds

4 0
3 years ago
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Sa = 2pi r h + 2pi r²
= 1327.32 units squared
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