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AVprozaik [17]
2 years ago
12

A proper unbiased coin was tossed 10 times for 3 trials, giving TTHHTHTTHH, TTTTTHHHHH, and THTHHHTTHT (T = Tails; H = Heads). W

hat is the difference between the theoretical and experimental probabilities of getting heads?
Mathematics
2 answers:
Flura [38]2 years ago
4 0
<h2>Answer:</h2>

The difference between the theoretical and experimental probabilities of getting heads is:

                                      0

<h2>Step-by-step explanation:</h2>
  • We know that while tossing a coin there are just two outcomes either a head will turn up or a tail.

Hence, there are 50-50 chance of getting a head or a tail.

Hence, the theoretical probability is: 1/2

  • For finding the experimental probability

The <em>first trial</em> is:

               TTHHTHTTHH

In this trial out of 10 times the head turned up 5 times.

Hence, the probability of head is: 5/10=1/2

The <em>second trial</em> is:

            TTTTTHHHHH

Here also out of 10 times the head turned up 5 times.

Hence, the probability of getting head is: 5/10=1/2

The <em>third trial</em> is:

                  THTHHHTTHT

In this trial too the head turned up 5 times out of 10 times.

Hence, the probability of getting head is: 5/10=1/2

Hence, we notices that in each of the trials the probability is: 1/2

i.e. The experimental probability of getting a head is: 1/2

Hence, the difference between the theoretical and experimental probability is:

                 1/2-1/2=0

Rashid [163]2 years ago
3 0
Theoretical is what you think will be the outcome vs. actually doing it...hope this helps
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a sports statistician claim that the mean winning times for boston marathon women's open division champions is at least 2.68 hou
GenaCL600 [577]

Answer:

We conclude that the mean winning times for Boston marathon women's open division champions is at least 2.68 hours.

Step-by-step explanation:

We are given that a sports statistician claim that the mean winning times for Boston marathon women's open division champions is at least 2.68 hours.

The mean winning time of a sample of 30 randomly selected Boston marathon women's open division champions is 2.60 hours. assume the population standard deviation is 0.32 hours.

<em>Let </em>\mu<em> = </em><u><em>mean winning times for Boston marathon women's open division champions.</em></u>

So, Null Hypothesis, H_0 : \mu\geq 2.68 hours      {means that the mean winning times for Boston marathon women's open division champions is at least 2.68 hours}

Alternate Hypothesis, H_A : p < 2.68 hours      {means that the mean winning times for Boston marathon women's open division champions is less than 2.68 hours}

The test statistics that would be used here <u>One-sample z test</u> <u>statistics</u> as we know about the population standard deviation;

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

where, \bar X = sample mean winning time = 2.60 hours

            \sigma = population standard deviation = 0.32 hours

            n = sample of women's open division champions = 30

So, <em><u>test statistics</u></em>  =   \frac{2.60-2.68}{\frac{0.32}{\sqrt{30} } }

                               =  -1.37

The value of z test statistics is -1.37.

Now, P-value of the test statistics is given by following formula;

         P-value = P(Z < -1.37) = 1 - P(Z \leq 1.37)

                       = 1 - 0.9147 = <u>0.0853</u>

Since, P-value of the test statistics is more than the level of significance as 0.0853 > 0.05, so <u>we have insufficient evidence to reject our null hypothesis</u> as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the mean winning times for Boston marathon women's open division champions is at least 2.68 hours.

6 0
2 years ago
a dealer bought some tires for 6500 the tires were sold for 9500 making $50 on each tire how many tires were involved
torisob [31]
130 tires were involved.
4 0
2 years ago
Read 2 more answers
James has to fill 40 water bottles for the soccer team. Each bottle holds 500 milliliters of water. How many liters of water doe
zhenek [66]

Answer:

20 litter

Step-by-step explanation:

40 \: bottels \:  \times 500 \times  ({10}^{ - 3} ) \: litters \:  = 20 \: litters

5 0
2 years ago
Simplify!!! <br><br>I need help please ​
satela [25.4K]

Answer:

\frac{4^{21}}{5^{6}}

Step-by-step explanation:

\frac{4^{7*3}}{5^{2*3}}=\frac{4^{21}}{5^{6}}

3 0
2 years ago
A lake contains 4 distinct types of fish. Suppose that each fish caught is equally likely to be any one of these types. Let Y de
strojnjashka [21]

Answer:

a) P(μ-k*σ≤ Y ≤ μ+k*σ ) ≥ 0.90

a= μ-3.16*σ , b= μ+3.16*σ

b) P(Y≥ μ+3*σ ) ≥ 0.90

b= μ+3*σ

Step-by-step explanation:

from Chebyshev's inequality for Y

P(| Y - μ|≤ k*σ ) ≥ 1-1/k²

where

Y =  the number of fish that need be caught to obtain at least one of each type

μ = expected value of Y

σ = standard deviation of Y

P(| Y - μ|≤ k*σ ) = probability that Y is within k standard deviations from the mean

k= parameter

thus for

P(| Y - μ|≤ k*σ ) ≥ 1-1/k²

P{a≤Y≤b} ≥ 0.90 →  1-1/k² = 0.90 → k = 3.16

then

P(μ-k*σ≤ Y ≤ μ+k*σ ) ≥ 0.90

using one-sided Chebyshev inequality (Cantelli's inequality)

P(Y- μ≥ λ) ≥ 1- σ²/(σ²+λ²)

P{Y≥b} ≥ 0.90  →  1- σ²/(σ²+λ²)=  1- 1/(1+(λ/σ)²)=0.90 → 3= λ/σ → λ= 3*σ

then for

P(Y≥ μ+3*σ ) ≥ 0.90

5 0
2 years ago
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