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Salsk061 [2.6K]
4 years ago
9

10 less than the product of 8 and 5

Mathematics
2 answers:
gulaghasi [49]4 years ago
4 0
The answer to this question is 30
Nastasia [14]4 years ago
3 0
The answer is 30.

8 x 5 =40

40 - 10 = 30
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3 years ago
Suppose that the lifetime of tv tubes are normally distributed with a standard deviation of 1.1 years. Suppose that exactly 20%
IrinaK [193]

Answer:

4.924 years

Step-by-step explanation:

Lets denote X the lifetime of a tv tube (In years). X has distribution N(\mu, 1.1) , with \mu unknown.

We know that P(X < 4) = 0.2. Using this data, we can find the value of \mu throught standarization.

Lets call Z = \frac{X - \mu}{1.1} the standarization of X. Z has distribution N(0,1), and its cummulative function, \phi is tabulated. The values of \phi can be found in the attached file.

P(X < 4) = P(\frac{X - \mu}{1.1} < \frac{4 - \mu}{1.1}) = P(Z < \frac{4 - \mu}{1.1}) = \phi(\frac{4 - \mu}{1.1}) = 0.2

The value q such that \phi (q) = 0.2 doesnt appear on the table. We can find it by using the symmetry of the normal density function. The opposite of q, -q must verify that \phi(-q) = 0.8 , hence -q must be equal to 0.84. Thus, q = -0.84

But this value of q should match with the number \frac{4 - \mu}{1.1} , so we have

\frac{4- \mu}{1.1} = -0.84

4 - \mu = 1.1 * (-0.84) = -0.924

\mu = 4 - (-0.924) = 4.924

Thus, the expected lifetime of TV tubes is 4.924 years.

I hope this works for you!

Download pdf
7 0
4 years ago
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