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evablogger [386]
3 years ago
9

Denis examined the records of the clients who had joined his gym more than six months ago, and he found these statistics: \begin

{aligned} &P(\text{joined in January})=0.12 \\\\ &P(\text{member for over 6 months})=0.5 \\\\ &P(\text{over 6 months and January})=0.024 \end{aligned} ​ P(joined in January)=0.12 P(member for over 6 months)=0.5 P(over 6 months and January)=0.024 ​ Find the probability that a client remained a member for more than 666 months, given that the client joined in January.
Mathematics
1 answer:
drek231 [11]3 years ago
8 0

Answer:

.2

Step-by-step explanation:

This is a conditional probability question. So it's asking you to find the probability that a client remained a member for more than 6 months, given that the client joined in January - which is formatted as P = (.5 | .12). You would then divide the chance of being over 6 months AND in January over the chance of being a member for over six months. ( .024 / .12) There, you would get .2 as your answer.

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Answer:

Option A

Step-by-step explanation:

Given information:

n_1=10, n_2=10

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Using the given information we get

p_1=\frac{x_1}{n_1}=\frac{3}{10}=0.3

p_2=\frac{x_2}{n_2}=\frac{5}{10}=0.5

The formula for confidence interval for the difference of proportions is

C.I.=(p_1-p_2)\pm z*\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}

From the standard normal table the value of z* at 5% confidence interval is 1.96.

Substitute the given values in the above formula.

C.I.=(0.3-0.5)\pm z*\sqrt{\frac{0.3(1-0.3)}{10}+\frac{0.5(1-0.5)}{10}}

C.I.=-0.2\pm (1.96)\sqrt{0.021+0.025}

C.I.=-0.2\pm (1.96)\sqrt{0.046}

C.I.=-0.2\pm 0.42

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Therefore, the correct option is A.

6 0
3 years ago
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