The number of pieces of popcorn in a large movie theater popcorn bucket is normally distributed, with a mean of 1610 and a stand
ard deviation of 10. Approximately what percentage of buckets contain between 1600 and 1620 pieces of popcorn?
A. approximately 68%
B. approximately 75%
C. approximately 95%
D. approximately 99.7%
A. approximately 68%
B. approximately 75%
C. approximately 95%
D. approximately 99.7%
2 answers:
Answer:
A. Approximately 68%
Step-by-step explanation:
Number of pieces of popcorn in a bucket is normally distributed.
Mean = μ = 1610
Standard deviation = б = 10
We know that in normal distribution, approximately 34% of bags will fall with in one standard deviation on one side. On both sides within the range of 1 standard deviation, 34 + 34 = 68 % of bags will fall.
A table of normal distribution about mean is attached for better understanding.
Our range is:
1600 to 1620
1610 - 10 to 1610 + 10
μ ± б
mean ± 1 standard deviation
Our range is within 1 standard deviation. Hence, Approximately 68% of buckets contain between 1600 and 1620 pieces of popcorn.
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Answer:
1. Mean square B= 5.32
2. Mean square E= 16.067
3. F= 0.33
4. p-value: 0.28
Step-by-step explanation:
Hello!
You have the information of 3 groups of people.
Group 1
n₁= 20
X[bar]₁= 3.2
S₁²= 14.3
Group 2
n₂= 20
X[bar]₂= 4.2
S₂²= 17.2
Group 3
n₃= 20
X[bar]₃= 7.6
S₃²= 16.7
1. To manually calculate the mean square between the groups you have to calculate the sum of square between conditions and divide it by the degrees of freedom.
Df B= k-1 = 3-1= 2
Sum Square B is:
∑ni(Ÿi - Ÿ..)²
Ÿi= sample mean of sample i ∀ i= 1,2,3
Ÿ..= general mean is the mean that results of all the groups together.
General mean:
Ÿ..= (Ÿ₁ + Ÿ₂ + Ÿ₃)/ 3 = (3.2+4.2+7.6)/3 = 5
Sum Square B (Ÿ₁ - Ÿ..)² + (Ÿ₂ - Ÿ..)² + (Ÿ₃ - Ÿ..)²= (3.2 - 5)² + (4.2 - 5)² + (7.6 - 5)²= 10.64
Mean square B= Sum Square B/Df B= 10.64/2= 5.32
2. The mean square error (MSE) is the estimation of the variance error (σ →
), you have to use the following formula:
Se²=<u> (n₁-1)S₁² + -(n₂-1)S₂² + (n₃-1)S₃²</u>
n₁+n₂+n₃-k
Se²=<u> 19*14.3 + 19*17.2 + 19*16.7 </u>= <u> 915.8 </u> = 16.067
20+20+20-3 57
DfE= N-k = 60-3= 57
3. To calculate the value of the statistic you have to divide the MSB by MSE
4. P(F ≤ F) = P(F
≤ 0.33) = 0.28
I hope you have a SUPER day!