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serg [7]
3 years ago
9

Skyler buys a bag of cookies that contains 9 chocolate chip cookies, 5 peanut butter cookies, 9 sugar cookies and 7 oatmeal rais

in cookies. What is the probability that Skyler randomly selects an oatmeal raisin cookie from the bag, eats it, then randomly selects another oatmeal raisin cookie?
Mathematics
1 answer:
sergiy2304 [10]3 years ago
7 0

Answer:   \dfrac{7}{145} .

Step-by-step explanation:

Given ,The bag contains :

9 chocolate chip cookies, 5 peanut butter cookies, 9 sugar cookies and 7 oatmeal raisin cookies.

Total cookies = 9+5+9+7=30

Now , the number of ways to choose any two cookies ( one after another) :

^{30}C_2=\dfrac{30!}{2!(30-2)!}=\dfrac{30\times29\times28!}{2\times28!}=435

The number of way to choose two oatmeal raisins ( one after another) :

^{7}C_2=\dfrac{7!}{2!(7-2)!}=\dfrac{7\times6\times5!}{2\times5!}=21

Now , the probability that Skyler randomly selects an oatmeal raisin cookie from the bag, eats it, then randomly selects another oatmeal raisin cookie will be :

\dfrac{21}{435}=\dfrac{7}{145}

Hence, the required probability is \dfrac{7}{145} .

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A 95% confidence interval estimate of the mean number of hours a legal professional works on a typical workday is [7.44 hours, 10.56 hours].

Step-by-step explanation:

We are given that x is normally distributed with a known standard deviation of 12.6.

A sample of 250 legal professionals was surveyed, and the sample's mean response was 9 hours.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                               P.Q.  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample average mean response = 9 hours

            \sigma  = population standard deviation = 12.6

            n = sample of legal professionals = 250

            \mu = mean number of hours a legal professional works

<em>Here for constructing a 95% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.</em>

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                   of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.96) = 0.95

P( -1.96 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < -1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

P( \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ]

                                       = [ 9-1.96 \times {\frac{12.6}{\sqrt{250} } } , 9+1.96 \times {\frac{12.6}{\sqrt{250} } } ]

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Therefore, a 95% confidence interval estimate of the mean number of hours a legal professional works on a typical workday is [7.44 hours, 10.56 hours].

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