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Studentka2010 [4]

3 years ago

5

Suppose we roll a fair six-sided die and sum the values obtained on each roll, stopping once our sum exceeds 285. Approximate th

e probability that at least 75 rolls are needed to get this sum. Probability

1 answer:

OlgaM077 [116]3 years ago

4 0

Answer:

1

Step-by-step explanation:

Probability is given by number of possible outcomes ÷ number of total outcomes

Assuming we stop rolling the six-sided die once our sum is 290 ( exceeds 285)

Number of possible outcomes = 75, number of total outcomes = 290

Probability (75 rolls are needed to get this sum) = 75/290 = 0.259

Probability (more than 75 rolls are needed to get this sum) = 1 - 0.259 = 0.741

Probability (at least 75 rolls are needed to get this sum) means that either 75 rolls or more than 75 rolls are needed to get this sum = 0.259 + 0.741 = 1

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kvasek [131]

Answer:

$t=\frac{85.8-113.4}{\sqrt{\frac{(50)^2}{145}+\frac{(75)^2}{141}}}}=-3.65$

$p_v =2*P(t_{(284)}$

Using Excel we can use the following code: "=2*T.DIST(-3.65;284;TRUE)"

So the p value is a very low value and using any significance level giveen $\alpha=0.05$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the means are different.

Step-by-step explanation:

1) Data given and notation

$\bar X_{M}=85.8$ represent the mean for the males

$\bar X_{F}=113.4$ represent the mean for females

$s_{M}=50$ represent the sample standard deviation for males

$s_{F}=75$ represent the sample standard deviation for female

$n_{M}=145$ sample size for the group poisoned

$n_{F}=141$ sample size for the group unpoisoned

t would represent the statistic (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for the two groups are equal or not , the system of hypothesis would be:

Null hypothesis: $\mu_{M} = \mu_{F}$

Alternative hypothesis: $\mu_{M} \neq \mu_{F}$

If we analyze the size for the samples both are higher than 30 but the population deviations are not given, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{M}-\bar X_{F}}{\sqrt{\frac{s^2_{M}}{n_{M}}+\frac{s^2_{F}}{n_{F}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

3) Calculate the statistic

We can replace in formula (1) the results obtained like this:

$t=\frac{85.8-113.4}{\sqrt{\frac{(50)^2}{145}+\frac{(75)^2}{141}}}}=-3.65$

4) Statistical decision

The first step is calculate the degrees of freedom, on this case:

$df=n_{M}+n_{F}-2=145+141-2=284$

Since is a two tailed test the p value would be:

$p_v =2*P(t_{(284)}$

Using Excel we can use the following code: "=2*T.DIST(-3.65;284;TRUE)"

So the p value is a very low value and using any significance level giveen $\alpha=0.05$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the means are different.

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3 years ago

The temperature function (in degrees Fahrenheit) in a three dimensional space is given by T(x, y, z) = 3x + 6y - 6z + 1. A bee i

madam [21]

You're looking for the extreme values of $T(x,y,z)=3x+6y-6z+1$ subject to $x^2+y^2+z^2=9$ . The Lagrangian is

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$L_z=-6+2\lambda z=0\implies z=\dfrac3\lambda$

$L_\lambda=x^2+y^2+z^2-9=0$

Substituting the first three solutions into the last equation gives

$\dfrac9{4\lambda^2}+\dfrac9{\lambda^2}+\dfrac9{\lambda^2}=9$

$\implies\lambda=\pm\dfrac32$

$\implies x=1,y=2,z=-2\text{ or }x=-1,y=-2,z=2$

At these points, we have

$T(1,2,-2)=28$

$T(-1,-2,2)=-26$

so the highest temperature the bee can experience is 28º F at the point (1, 2, -2), and the lowest is -26º F at the point (-1, -2, 2).

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