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zvonat [6]
2 years ago
10

I need help on this question

Mathematics
2 answers:
olga2289 [7]2 years ago
8 0

Hey there!

-1 ¼ • 9

= - (4+1)/4 • 9

= -5/4 • 9

= -(5 × 9)/4

= <u>-</u><u> </u><u>4</u><u>5/4</u>

= <u>- 11.25</u>

Hope it helps ya!

EastWind [94]2 years ago
6 0

-11 1/4

hope this was helpful :)

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Answer:

the probability is 0.0750

Step-by-step explanation:

The computation of the probability is shown below;

The mean of x is

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Now the probability is

= P(Z < -0.3 ÷ 0.1688) + P(\bar{x} > 0.3 ÷ 0.1688)

= P(Z < -1.78) + P(Z > 1.78)

= P(Z < -1.78) + 1 - P(Z < 1.78

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The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control a
Dafna11 [192]

Answer:

Probability that at least 490 do not result in birth defects = 0.1076

Step-by-step explanation:

Given - The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC). A local hospital randomly selects five births and lets the random variable X count the number not resulting in a defect. Assume the births are independent.

To find - If 500 births were observed rather than only 5, what is the approximate probability that at least 490 do not result in birth defects

Proof -

Given that,

P(birth that result in a birth defect) = 1/33

P(birth that not result in a birth defect) = 1 - 1/33 = 32/33

Now,

Given that, n = 500

X = Number of birth that does not result in birth defects

Now,

P(X ≥ 490) = \sum\limits^{500}_{x=490} {^{500} C_{x} } (\frac{32}{33} )^{x} (\frac{1}{33} )^{500-x}

                 = {^{500} C_{490} } (\frac{32}{33} )^{490} (\frac{1}{33} )^{500-490}  + .......+ {^{500} C_{500} } (\frac{32}{33} )^{500} (\frac{1}{33} )^{500-500}

                = 0.04541 + ......+0.0000002079

                = 0.1076

⇒Probability that at least 490 do not result in birth defects = 0.1076

4 0
2 years ago
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