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MAT 120 Final Exam Practice Test Questions With Well Elaborated Answers A+ Score.
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The mean value of land and buildings per acre from a sample of farms is $1300, with a standard deviation of $100. The data set has a bell-shaped distribution. Assume the number of farms in the sample is 75. (a) Use the empirical rule to estimate the number of farms whose land and building val...

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MAT 120 Final Exam Practice Test

The mean value of land and buildings per acre from a sample of farms is $1300,

with a standard deviation of

$100. The data set has a bell-shaped distribution. Assume the number of farms in the sample is 75.



(a) Use the empirical rule to estimate the number of farms whose land and building values per acre are
between

$1200 and $1400.

___ farms.



(b) If 26 additional farms were sampled, about how many of these additional farms would you expect to
have land and building values between

$1200 per acre and $1400 per acre?

___ farms out of 336. - correct answer Section 2.4



ANSWER:

(a) 51.



(b) 18.




Work:

(a)

n = 75

mean = 1300.

standard deviations = 100

,between 1200 and 1300 is 100. One standard deviation.

Between 1400 and 1300 is 100. One standard deviation.



68% of data falls within 1 standard deviation.



75 * 0.68 = 51.




(b)

26 * 0.68 = 17.68. round to 18.



The monthly utility bills in a city are normally distributed, with a mean of $100 and a standard deviation
of $12.

Find the probability that a randomly selected utility bill is (a) less than $67, (b) between $83 and $100,
and (c) more than $130.



(a) The probability that a randomly selected utility bill is less than $67 is ___?



(b) The probability that a randomly selected utility bill is between $83 and $100 is ___?



(c) The probability that a randomly selected utility bill is more than $130 is __? - correct answer
Section 5.2



ANSWER:

(a) 0.0030.



(b) 0.4222.



(c) 0.0062

, Work:

(a) (67 - 100)/12 = -2.75.... -2.75 ON CHART 2. is 0.0030.



(b)

(83-100)/12 = -1.42.... USE CHART 2.

-1.42 on chart is 0.0778.

(100-100)/13 = 0.... 0 on chart is 0.5000.

0.5000 - 0.0778 = 0.4222



(c)

(130-100)/12 = 2.5.... 2.5 on chart is 0.9938.

1 - 0.9938 = 0.0062.



The data given below show the number of overtime hours worked in one week per employee. Use the
data to complete parts (a) and (b).



Overtime hours | Employees

0|8

1 | 12

2 | 30

3 | 53

4 | 42

5 | 26

6 | 19



(a) Construct the probability distribution by completing the table below.

x | P(x)

0|

1|

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