Five consecutive positive integers are chosen at random. If the average of the five integers is odd, what is the remainder when
1 answer:
Answer:
The remainder would be 1 or 3
Step-by-step explanation:
Consider the 5 consecutive positive integers are,
x, x + 1, x + 2, x + 3, x + 4,
Since, the average of these 5 numbers =
= x + 2
If x + 2 = odd,
⇒ x = odd - 2 = odd - even = odd
⇒ x + 4 = odd + even = odd
∵ an odd number is represented by '2n + 1'
Where, n = 0, 1, 2, 3, ........
Now, 2n + 1 = 1( mod 4) if n = even
While, 2n + 1 = 3( mod 4) if n = odd,
Hence, when the largest of the five integers is divided by 4 remainder would be 1 or 3.
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Answer:
The p-value is less than the significance level, so we will reject the null hypothesis, and conclude that at 5% significance level, the proportion for women that ran 5 km is more than the proportion of male that ran the 5 km. Thus, women are more likely to run the 5 km race.
Step-by-step explanation:
We are given;
Number of males selected; n₁ = 150
Number of females selected; n₂ = 190
Proportion of the males; p₁ = 0.6
Proportion of females; p₂ = 0.5
Let's define the hypotheses;
Null hypothesis; H0: p₁ ≥ p₂
Alternative hypothesis; Ha: p₁ < p₂
Now, the z score formula for this is;
z = (p₁ - p₂)/√(p^(1 - p^))(1/n₁ + 1/n₂))
Where;
p^ = (p₁ + p₂)/2
p^ = (0.6 + 0.5)/2
p^ = 0.55
Thus;
z = (0.6 - 0.5)/√(0.55(1 - 0.55))((1/150) + (1/190))
z = 0.1/0.0543
z = 1.84
From online p-value from z-score calculator attached, using z = 1.84, significance level = 0.05, one tailed hypothesis, we have;
P-value = 0.033
The p-value is less than the significance level, so we will reject the null hypothesis, and conclude that at 5% significance level, the proportion for women that ran 5 km is more than the proportion of male that ran the 5 km. Thus, women are more likely to run the 5 km race.
Although if we check at 0.01 significance level, we will not have the same answer as the p-value will be greater.
