Answer:
Step-by-step explanation:
The PDF of X is
The PDF of Y is
The means of X and Y are respectively,
so we can see that the larger the parameter, the smaller the mean. Hence the PDF of Z = min(X, Y) is an exponential with the largest parameter of the two.
Therefore, the PDF of Z is
For large sample confidence intervals about the mean you have:
xBar ± z * sx / sqrt(n)
where xBar is the sample mean z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α sx is the sample standard deviation n is the sample size
We need only to concern ourselves with the error term of the CI, In order to find the sample size needed for a confidence interval of a given size.
z * sx / sqrt(n) = width.
so the z-score for the confidence interval of .98 is the value of z such that 0.01 is in each tail of the distribution. z = 2.326348
The equation we need to solve is:
z * sx / sqrt(n) = width
n = (z * sx / width) ^ 2.
n = ( 2.326348 * 6 / 3 ) ^ 2
n = 21.64758
Since n must be integer valued we need to take the ceiling of this solution.
n = 22
Answer:
Step-by-step explanation:
Total percentage = 100%
And we know that proportions of Vanilla +chocolate +strawberry = 100%
Proportion of Vanilla = 45%
Proportion of Strawberry = 1/4 *100 = 25%
Let proportion of chocolate be represented by variable C;
So, 45% +25% + C = 100%
65% +C = 100%
Subtract 65% from both sides to solve for C;
C=100% - 65%
C = 35%
Therefore there is 35% of chocolate in the shop.
