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vodka [1.7K]
3 years ago
6

NEED HELP!! Scenario #1: You flip a coin and roll a six sided die. What is the probability that you will flip heads and roll a n

umber greater than 2? Scenario #2: You spin a colored wheel and roll a six sided die. The colored wheel has 4 different colors (blue, green, red, and orange). What is the probability that you will spin on blue or roll a 6?
• What is the probability for the scenario that gives you the better odds? if you determine that the odds are the same, then document the probability for the either of scenarios.
Mathematics
1 answer:
leva [86]3 years ago
4 0

Scenario #1:

Flipping a coin and rolling a die are independent events, so the combined probability is the product of the individual probabilities.

Flipping a coin: There are two different possible outcomes, heads and tails. You are interested in heads. You are interested in 1 outcome out of 2.

p(heads) = 1/2

Rolling a die: There are 6 different possible outcomes, the numbers 1, 2, 3, 4, 5, and 6. You are interested in a number greater than 2, so your desired outcomes are 3, 4, 5, 6, which means 4 different desired outcomes out of 6.

p(die toss greater than 2) = 4/6 = 2/3

Combined probability:

p(heads followed by number greater than 2) =

= p(heads) * p(die toss greater than 2)

= 1/2 * 2/3

= 1/3

Scenario #2:

There are 4 colors on the spinner. For each color on the spinner, the die can land on one of 6 numbers. 4 * 6 = 24. There are 24 different combined outcomes.

You are interested in blue OR 6. Six outcomes have blue and a number from 1 to 6. Then if the spinner lands in any of the other 3 colors, each one has one outcome of 6 with that color. The total number of desired outcomes is 6 + 3 = 9.

Number of desired outcomes: 9

Number of possible outcomes: 24

p(blue or 6) = 9/24 = 3/8

Final Question:

The scenario that gives you better odds is Scenario #2 since 3/8 > 1/3.


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Answer:

Null hypothesis:\mu = 100  

Alternative hypothesis:\mu \neq 100  

t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921  

p_v =2*P(t_{11}  

Step-by-step explanation:

1) Data given and notation  

Data: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4

We can calculate the sample mean and deviation for this data with the following formulas:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}

The results obtained are:

\bar X=98.375 represent the sample mean  

s=0.6.109 represent the sample standard deviation  

n=12 sample size  

\mu_o =100 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 100pCL/L, the system of hypothesis are :  

Null hypothesis:\mu = 100  

Alternative hypothesis:\mu \neq 100  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

3) Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921  

4) P-value  

First we need to find the degrees of freedom for the statistic given by:

df=n-1=12-1=11

Since is a two sided test the p value would given by:  

p_v =2*P(t_{11}  

5) Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significant different from 100 at 5% of significance.  

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