A. 2/10 is the correct answer
Answer:
The probability that can afford to spend between $800 and $900
P(800≤X≤900) = 0.6826
The percentage of that can afford to spend between $800 and $900
P(800≤X≤900) = 68 percentage
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given that the mean of the Normal distribution = $850
Given that the standard deviation of the Normal distribution = $50
Let 'X' be a random variable in a normal distribution
Let x₁ = 800
Let x₂ =850
<u><em>Step(ii):-</em></u>
The probability that can afford to spend between $800 and $900
P(800≤X≤900) = P(-1≤Z≤1)
= P(Z≤1) - P(Z≤-1)
= 0.5 + A(1) - (0.5 - A(-1))
= A(1) +A(-1)
= 2× A(1) (∵ A(-1) =A(1)
= 2 × 0.3413
= 0.6826
The percentage of that can afford to spend between $800 and $900
P(800≤X≤900) = 68 percentage
Answer:
d. (0.737, 0.823)
The 90% confidence interval is = (0.737, 0.823)
Step-by-step explanation:
Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.
The confidence interval of a statistical data can be written as.
p+/-z√(p(1-p)/n)
Given that;
Proportion p = 195/250 = 0.78
Number of samples n = 250
Confidence interval = 90%
z value(at 90% confidence) = 1.645
Substituting the values we have;
0.78 +/- 1.645√(0.78(1-0.78)/250)
0.78 +/- 1.645√(0.0006864)
0.78 +/- 1.645(0.026199236630)
0.78 +/- 0.043097744256
0.78 +/- 0.043
(0.737, 0.823)
The 90% confidence interval is = (0.737, 0.823)
Answer: -1.5, -1, -1/2, 1/4
Step-by-step explanation:
Answer:
it will go in once you add
Step-by-step explanation: