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.
-10/8 and + 15/8. Those should work.
Data:
<span>Number of shares: 500
name of fund
nav offer price
hat mid-cap $18.94 $19.14
Profit $6,250.
1) Investment = number of shares * offer price = <em />500 * $ 19.14 = $ 9570
2) Total net value = profit + investment = $6250 + $9570 = $15820
3) NAV = total net value / number of shares = $15820 / 500 = $31.64
Answer: option c: 31.64
</span>
Im going to go on and say that it is 90
Answer:
2,1,0,-1,-2
Step-by-step explanation: