Answer:
Yes
Step-by-step explanation:
First, suppose that nothing has changed, and possibility p is still 0.56. It's our null hypothesis. Now, we've got Bernoulli distribution, but 30 is big enough to consider Gaussian distribution instead.
It has mean μ= np = 30×0.56=16.8
standard deviation s = √npq
sqrt(30×0.56×(1-0.56)) = 2.71
So 21 is (21-16.8)/2.71 = 1.5494 standard deviations above the mean. So the level increased with a ˜ 0.005 level of significance, and there is sufficient evidence.
Answer:
70
Step-by-step explanation:
7 and 10 = 70, This is the correct answer because 7 times 10 equals 70, and 10 times 7 equals 70 too, so 70 is the LCD
7 and 10 = 1, This is impossible because anything multiplied by 7 naturally does not get 1 as a answer, including 10 too. X
7 and 10 = 10, Only 10 times 1 equals 10, but 7 a natural whole number does not equal 10. X
7 and 10 = 35, Only 7 times 5 equals 35, but 10 times a whole number does not equal 35. X
Answer:
Probability that a randomly selected firm will earn less than 100 million dollars is 0.8413.
Step-by-step explanation:
We are given that the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 5 million dollars. Also, incomes for the industry are distributed normally.
<em>Let X = incomes for the industry</em>
So, X ~ N(
)
Now, the z score probability distribution is given by;
Z =
~ N(0,1)
where,
= mean income of firms in the industry = 95 million dollars
= standard deviation = 5 million dollars
So, probability that a randomly selected firm will earn less than 100 million dollars is given by = P(X < 100 million dollars)
P(X < 100) = P(
<
) = P(Z < 1) = 0.8413 {using z table]
Therefore, probability that a randomly selected firm will earn less than 100 million dollars is 0.8413.
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