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1 year ago

7

In a school of 330 students, 85 of them are in the drama club, 200 of them are in a sports team and 60 students do drama and spo

rts.
Find the probability that a student chosen at random isn’t in the drama club or in a sports team.

1 answer:

liq [111]1 year ago

7 0

Let's denote students in D = in the drama club , S<span> = in a sports team </span>
<span>P(D) = 85/330 </span>
<span>P(S) = 200/330 </span>
<span>
P(D and S) = 60/300 </span>
<span>
P(D or S) = P(D) + P(S) - P(D and S) </span><span>= 85/330 + 200/330 - 60/330 = 15/22 </span>
<span>
P(neither D nor S) = 1- 15/22 </span><span>= 7/22</span>

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Given that:

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$\text{P(green swordtail)}=\dfrac{\text{Total green swordtail fish}}{\text{Total fish}}$

$\text{P(green swordtail)}=\dfrac{80}{140}$

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B. We have to find the probability that the selected fish is male.

$\text{P(Male fish)}=\dfrac{\text{Total male fish}}{\text{Total fish}}$

$\text{P(Male fish)}=\dfrac{36+24}{140}$

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Therefore, the probability that the selected fish is a male, is $\dfrac{3}{7}.$

C. We have to find the probability that the selected fish is a male green swordtail.

$\text{P(Male green swordtail)}=\dfrac{\text{Total male green swordtail fish}}{\text{Total fish}}$

$\text{P(Male green swordtail)}=\dfrac{36}{140}$

$\text{P(Male green swordtail)}=\dfrac{9}{35}$

Therefore, probability that the selected fish is a male green swordtail is $\dfrac{9}{35}.$

D.

We have to find the probability that the selected fish is either a male or a green swordtail.

$\text{P(Male or green swordtail)}=\dfrac{\text{Total male or green swordtail fish}}{\text{Total fish}}$

$\text{P(Male or green swordtail)}=\dfrac{44+36+24}{140}$

$\text{P(Male or green swordtail)}=\dfrac{96}{140}$

$\text{P(Male or green swordtail)}=\dfrac{24}{35}$

Therefore, the probability the selected fish is either a male or a green swordtail is $\dfrac{24}{35}.$

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Thepotemich [5.8K]

<h2><u>Answer with explanation</u>:</h2>

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$t_c=t_{\alpha/2, df}=t_{0.05,\ 64}=1.9977$

Then , 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel would be :

$19.5\pm(1.9977)\dfrac{5.2}{\sqrt{65}}$

$=19.5\pm1.288$

$(19.5-1.288,\ 19.5+1.288)=(18.212,\ 20.788)$

∴ 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel : (18.212, 20.788)

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