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1 year ago

8

(b) Find z such that 39% of all observations from a standard normal distribution are greater than z. (Enter your answer rounded

to two decimal places.)

1 answer:

vova2212 [387]1 year ago

3 0

For 39% of all observations that are greater than z on the normal distribution table, the percentage of observations that are lesser than z is

100 - 39 = 61%

The proportion is 61/100 = 0.61

From the normal distribution table, the value of z corresponding to this proportion is 0.28

Thus, the required value of z is 0.28

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$\frac{2\pi}{15},\frac{4\pi}{15},\frac{8\pi}{15},\frac{2\pi}{3},\frac{14\pi}{15}, \frac{16\pi}{15}, \frac{4\pi}{3},\frac{22\pi}{15}, \frac{26\pi}{15}, \frac{28\pi}{15}$

Step-by-step explanation:

Solving trigonometric equations.

We are given a condition and we must find all angles who meet it in the provided interval. Our equation is

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To find all the solutions, we'll give n values of 0, 1, 2,... until x stops belonging to the interval $(0,2\pi)$

For n=0

$x=\frac{\frac{2\pi}{3}}{5}=\frac{2\pi}{15}$

$x=\frac{\frac{4\pi}{3}}{5}=\frac{4\pi}{15}$

For n=1

$x=\frac{\frac{2\pi}{3}+2\pi}{5}=\frac{8\pi}{15}$

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For n=2

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For n=3

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$x=\frac{\frac{4\pi}{3}+6\pi}{5}=\frac{22\pi}{15}$

For n=4

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$x=\frac{\frac{4\pi}{3}+8\pi}{5}=\frac{28\pi}{15}$

For n=5 we would find values such as

$x=\frac{\frac{2\pi}{3}+10\pi}{5}=\frac{32\pi}{15}$

$x=\frac{\frac{4\pi}{3}+10\pi}{5}=\frac{34\pi}{15}$

which don't lie in the interval $(0,2\pi)$

The whole set of results is

$\frac{2\pi}{15},\frac{4\pi}{15},\frac{8\pi}{15},\frac{2\pi}{3},\frac{14\pi}{15}, \frac{16\pi}{15}, \frac{4\pi}{3},\frac{22\pi}{15}, \frac{26\pi}{15}, \frac{28\pi}{15}$

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