Where is the rest of the question/answers????
Answer:
A) sample mean = $1.36 million
B) standard deviation = $0.9189 million
C) confidence interval = ($1.93 million , $0.79 million)
*since the sample size is very small, the confidence interval is not valid.
Step-by-step explanation:
samples:
- $2.7 million
- $2.4 million
- $2.2 million
- $2 million
- $1.5 million
- $1.5 million
- $0.5 million
- $0.5 million
- $0.2 million
- $0.1 million
sample mean = $1.36 million
the standard deviation:
- $2.7 million - $1.36 million = 1.34² = 1.7956
- $2.4 million - $1.36 million = 1.04² = 1.0816
- $2.2 million - $1.36 million = 0.84² = 0.7056
- $2 million - $1.36 million = 0.64² = 0.4096
- $1.5 million - $1.36 million = 0.14² = 0.0196
- $1.5 million - $1.36 million = 0.14² = 0.0196
- $0.5 million - $1.36 million = -0.86² = 0.7396
- $0.5 million - $1.36 million = -0.86² = 0.7396
- $0.2 million - $1.36 million = -1.16² = 1.3456
- $0.1 million - $1.36 million = -1.26² = 1.5876
- total $8.444 million / 10 = $0.8444 million
standard deviation = √0.8444 = 0.9189
95% confidence interval = mean +/- 1.96 standard deviations/√n:
$1.36 million + [(1.96 x $0.9189 million)/√10] = $1.36 million + $0.57 million = $1.93 million
$1.36 million - $0.57 million = $0.79 million
Approximately 2%. The reason is if you choose 20% and only have four with beef as the most popular, and approximately the number of restaurants is 15.
Answer:
y=x+4
Step-by-step explanation:
