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Sholpan [36]
3 years ago
8

Can you help me with this?

Mathematics
1 answer:
Firdavs [7]3 years ago
3 0
I don’t know this answer but I think it’s 81
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Read 2 more answers
Suppose the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 5 million dollars
GuDViN [60]

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(\mu=95,\sigma^{2}=5^{2})

Now, the z score probability distribution is given by;

         Z = \frac{X-\mu}{\sigma} ~ N(0,1)

where, \mu = mean income of firms in the industry = 95 million dollars

            \sigma = 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( \frac{X-\mu}{\sigma} < \frac{100-95}{5} ) = 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.

5 0
3 years ago
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