We can solve this problem by referring to the standard
probability distribution tables for z.
We are required to find for the number of samples given the
proportion (P = 5% = 0.05) and confidence level of 95%. This would give a value
of z equivalent to:
z = 1.96
Since the problem states that it should be within the true
proportion then p = 0.5
Now we can find for the sample size using the formula:
n = (z^2) p q /E^2
where,
<span> p = 0.5</span>
q = 1 – p = 0.5
E = estimate of 5% = 0.05
Substituting:
n = (1.96^2) 0.5 * 0.5 / 0.05^2
n = 384.16
<span>Around 385students are required.</span>
Answer:
(A) 0.125 probability
(B) 0.625 probability
(C) 660 miles
Step-by-step explanation:
The distance driven by a truck driver daily, falls between 300miles and 700miles and follows a uniform distribution.
(A) The probability that the truck driver goes more than 650 miles a day is:
[700 - 650] / [700 - 300] = 50/400 = 0.125
(B) The probability that the truck driver goes between 400 and 650 miles a day is:
[650 - 400] / [700 - 300] = 250/400 = 0.625
(C) The minimum number of miles the truck driver travels on the furthest 10% of days is given thus:
10% of 400 = 40
Subtract this from the farthest distance;
700miles - 40miles = 660miles
The answer to this equation would be 3.
Answer:
x = -5, y= 7. You can check this yourself by plugging in the numbers.
Step-by-step explanation:
I thinks is A or B if not I’m sorry
