Mathematics College

according to wine-searcher, wine critics generally use a wine-scoring scale to communicate their opinions on the relative quality of wines. wine scores range from to , with a score of indicating a great wine, indicating an outstanding wine, indicating a very good wine, indicating a good wine, indicating a mediocre wine, and below indicating that the wine is not recommended. random ratings of a pinot noir recently produced by a newly established vineyard in follow: excel file: data07-11.xlsx 87 91 86 82 72 91 60 77 80 79 83 96 a. develop a point estimate of mean wine score for this pinot noir (to decimals). 82.00 b. develop a point estimate of the standard deviation for wine scores received by this pinot noir (to decimals). 9.6389

Answers

Answer 1

(a) The point estimate of mean wine score for this pinot noir is 82.

(b) The point estimate of standard deviation wine score for this pinot noir is 9.6389.

Below table showing calculation of Point Estimate of Mean and Standard Deviation:

Score                                  X-X’                                  (X-X’)^2

87                                         5                                         25

91                                         9                                         81

86                                         4                                         16

82                                         0                                         0

72                                        -10                                       100

91                                         9                                         81

60                                        -22                                       484

77                                        -5                                         25

80                                        -2                                         4

79                                        -3                                         9

83                                         1                                         1

96                                         14                                       196

984                                                                                   1022

Mean(X’) = Total Score/n

n = Total number = 12

X’ = 984/12 = 82

Standard Deviation (σ) = √∑(X-X’)^2/(n-1)

σ = √1022/(12-1)

σ = √1022/ 11

σ = 9.6389

(a) The point estimate of mean wine score for this pinot noir is 82.

(b) The point estimate of standard deviation wine score for this pinot noir is 9.6389

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Related Questions

1. Which equation describes the line with
slope -4 and y-intercept 2?
A y=-4x+2
B y=-4x-2
C y=4x-2
D y = 4x + 2

Answers

Answer:

Step-by-step explanation:

The slope-intercept form of a linear equation is y = mx + b, where m is the slope and b is the y-intercept.

Therefore, the equation of the line with slope -4 and y-intercept 2 is y = (-4)x + 2.

A train travels at 80 miles per hour. An equation can be written that compares the time (t) with the distance (d). What is the domain and range?

1. The domain is distance (d) and the range is time (t).

2. The domain is time (t) and the range is distance (d).

3. The domain is time (t) and the range is 80.

4. The domain is 80 and the range is time (t).

Answers

The required answer is the domain is time (t) and the range is a distance (d) i.e. Option 2.

What are domain and range?

The value range that can be plugged into a function is known as its domain. In a function like f, this set represents the x values f(x). The collection of values that a function can take on is known as its range. The values that the function outputs when we enter an x value are in this set.

From the given question, and the above definition of domain and range,

the time (t) acts as an x-values or input value and the distance (d) acts as a y-value or output value

Hence, the domain is time (t) and the range is a distance (d) i.e. Option 2.

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There are currently 25 frogs in a (large) pond. The frog population grows exponentially, tripling every 10 days.
How long will it take (in days) for there to be 150 frogs in the pond?
Time to 150 frogs: days
The pond's ecosystem can support 1400 frogs. How long until the situation becomes critical?
Time to 1400 frogs: days

Answers

There are 21 days for there to be 150 frogs in the pond.

There are 37 days for there to be 1400 frogs in the pond.

What is exponential growth?

Quantity increases over time through a process called exponential growth. When a quantity's instantaneous rate of change with respect to time is proportional to the quantity itself, it happens.

Given:

There are currently 25 frogs in a (large) pond. The frog population grows exponentially, tripling every 10 days.

The exponential equation for the given problem is,

[tex]A(t) = Ar^t^/^1^0[/tex]

To find the number of days for there to be 150 frogs in the pond.

Here,

A(t) = 250, A = 25, r = 3

⇒

[tex]250 = 25(3)^t^/^1^0\\10 = 3^t^/^1^0\\t = 20.97[/tex]

t ≈ 21

Hence, there are 21 days for there to be 150 frogs in the pond.

Now to find how long for there to be 1400 frogs in the pond, we solve:

[tex]1400 = 25(3)^t^/^1^0\\56 = (3)^t^/^1^0\\t = 36.64[/tex]

t ≈ 37

Hence, there are 37 days for there to be 1400 frogs in the pond.

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