Question

According to a certain website, wine critics generally use a wine-scoring scale to communicate their opinions on the relative quality of wines. Wine scores range from 0 to 100, with a score of 95–100 indicating a great wine, 90–94 indicating an outstanding wine, 85–89 indicating a very good wine, 80–84 indicating a good wine, 75–79 indicating a mediocre wine, and below 75 indicating that the wine is not recommended. Random ratings of a pinot noir recently produced by a newly established vineyard in 2018 follow:
87 91 86 82 72 91
60 77 80 79 83 96

Required:
a. Develop a point estimate of mean wine score for this pinot noir.
b. Develop a point estimate of the standard deviation for wine scores received by this pinot noir. (Round your answer to four decimal places.)

Answer 1

Answer:

$\bar x = 82$

$\sigma = 9.64$

Step-by-step explanation:

Given

$n = 12$

$87\ 91\ 86\ 82\ 72\ 91\ 60\ 77\ 80\ 79\ 83\ 96$

Solving (a); Point estimate of mean

To do this, we simply calculate the sample mean

$\bar x = \frac{\sum x}{n}$

$\bar x = \frac{87+ 91+ 86+ 82+72+91+60+77+80+79+83+96}{12}$

$\bar x = \frac{984}{12}$

$\bar x = 82$

Solving (b); Point estimate of standard deviation

To do this, we simply calculate the sample standard deviation

$\sigma = \sqrt{\frac{\sum(x-\bar x)^2}{n - 1}$

$\sigma = \sqrt\frac{(87-82)^2+ (91-82)^2+ ....+ (79-82)^2+ (83-82)^2+ (96-82)^2}{12-1}$

$\sigma = \sqrt\frac{1022}{11}$

$\sigma = \sqrt{92.91}$

$\sigma = 9.64$

Note that: The sample mean and the sample standard deviation are the best point estimators for the mean and the standard deviation, respectively.

Hence, the need to solve for sample mean and sample standard deviation