Answer:
b. The values X and Y are independent therefore, the mean is 34 seconds and the standard deviation is 50 seconds
Step-by-step explanation:
Subtraction between normal variables:
When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.
Before the training, the mean running time for the students to run a mile was 402 seconds with standard deviation 40 seconds.
This means that 
After completing the program, the mean running time for the students to run a mile was 368 seconds with standard deviation 30 seconds.
This means that 
Which of the following is true about the distribution of X-Y?
They are independent, so:
This means that the correct answer is given by option b.
Part 1:
Given a number line with the point
and the point
The sampling error is given by:
Part 2:
Given a number line with the point
and the point
The sampling error is given by:
I would go with A but I am not 100% sure so you may want to check however you can.
Given
jacinta buys 4 pounds of turkey and 2 pounds of ham & pays a total of $30, turkey costs $1.50 less per pound than the ham.
find out the combined cost of 1 pound of turkey and 1 pound of ham
To proof
As given in the question
jacinta buys 4 pounds of turkey and 2 pounds of ham
total pay by jacinta = $30
let us assume that the price of the ham = x
as given in the question
turkey costs $1.50 less per pound than the ham
turkey costs becomes = x - 1.50
then the equation becomes
30 = 4 (x - 1.50) + 2x
30 = 4x + 2x - 6
36 = 6x
x = 6
thus
ham cost per pound = $6
turkey cost per pound = 6 - 1.50
= $ 4.5
Now find out
cost of 1 pound of turkey + 1 pound of ham = $ 6 + $ 4.5
= $ 10.5
Hence proved
Answer:
a) Calculate the probability that at least one of them suffers from arachnophobia.
x = number of students suffering from arachnophobia
= P(x ≥ 1)
= 1 - P(x = 0)
= 1 - [0.05⁰ x (1 - 0.05)¹¹⁻⁰
]
= 1 - (0.95)¹¹
= 0.4311999 = 0.4312
b) Calculate the probability that exactly 2 of them suffer from arachnophobia? 0.08666
= P(x = 2)
= (¹¹₂) x (0.05)² x (0.95)⁹
where ¹¹₂ = 11! / (2!9!) = (11 x 10) / (2 x 1) = 55
= 55 x 0.0025 x 0.630249409 = 0.086659293 = 0.0867
c) Calculate the probability that at most 1 of them suffers from arachnophobia?
P(x ≤ 1)
= P(x = 0) + P(x = 1)
= [(¹¹₀) x 0.05⁰ x 0.95¹¹] + [(¹¹₁) x 0.05¹ x 0.95¹⁰]
= (1 x 1 x 0.5688) + (11 x 0.05 x 0.598736939) = 0.5688 + 0.3293 = 0.8981
