Response with clarification:
Let p denote the proportion of adults in the town who have encountered this flu strain.
According to the provided information
∵ 
this is a two-tailed test.
Test statistic:
, where p= denotes the population proportion
= signifies the sample proportion
n= represents the sample size
Setting n= 6 and 
and p=0.08
P-value for the two-tailed test:[2P(Z>|z|)
=2P(Z>|-0.415|)
=2P(Z>0.415) = 2[1-P(Z≤0.415)] [∵ P(Z>z)=1-P(Z≤z)]
=2(1-0.6609) [from the z-table]
=0.6782
Decision: Because the p-value(0.6782) exceeds the significance level of 0.01, we do not reject the null hypothesis.
This leads us to conclude that there is insufficient evidence to back the assertion that the percentage of all adults in this town exposed to this flu strain deviates from the national average of 8%.
Response:
MAD value comes out to be 3.
Detailed Breakdown:
The given sales forecasts for the last four months are 5, 6, 11, and 12 units.
To calculate the Mean Absolute Deviation (MAD) for these forecasts:
The average of the forecasts across four months is 
.
Thus, the total of absolute differences between the forecast values and the average is = |5 - 8.5| + |6 - 8.5| + |11 - 8.5| + |12 - 8.5| = 3.5 + 2.5 + 2.5 + 3.5 = 12.
Hence, the MAD value will be = 
(Final Answer)
This problem can be addressed by applying the normal approximation to a binomial distribution.
Calculations:
Mean (μ) = np = 10,000 × 0.5 = 5,000
The standard deviation (σ) is given by:
The probability of obtaining more than 5,100 tails is 0.0228, whereas the probability of fewer than 5,100 tails is 0.9772.
Thus, the odds of having more than 5,100 tails are:
0.0228 : 0.9772 = 1 : 42.86.
