Changing improper fractions to mixed
form or mixed numbers is a very simple thing. You just have to follow the following:
Then write
Let us have 25/8.
<span><span>
1.
</span><span>Divide the numerator (top
number) by the denominator (number below
the bar). </span></span>
Simply divide 25 by 8.
25 ÷ 8 = n
We can have 3 and 1 as a remainder. So, the
answer will be 3 and 1/8.
<span><span>
2.
</span> <span>Write down the whole number answer.
Since we have 3 as the whole number answer,
write it down. And;</span></span>
<span><span>
3.
</span><span> Write down any remainder above the denominator,
like this:</span></span>
3 and 1 (remainder) over 8
(denominator of the given improper fraction).
That is, 3 1/8
Answer:
16
Step-by-step explanation:
The most important factor is the total budget $1816.50, for the soccer tournament.
1. Subtract $880.50 from $1816.50 = $936
2. We have $936 left, and a budget of $58.50 per player for meals
3. Divide 936/58.50 = 16
4. 16 soccer players can be brought to the tournament
It would take 16.875 gallons to fill the gasoline tank
Answer:
a) P(Y=10)=0.0013
b) P(Y≤5)=0.00000035
c) Mean = 17.5
S.D. = 2.29
Step-by-step explanation:
We can model this as a binomial random variable with n=25 and p=0.7.
The probability that k students from the sample are foreign students can be calculated as:
![P(y=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(y=k) = \dbinom{25}{k} 0.7^{k} 0.3^{25-k}\\\\\\](https://tex.z-dn.net/?f=P%28y%3Dk%29%20%3D%20%5Cdbinom%7Bn%7D%7Bk%7D%20p%5E%7Bk%7D%281-p%29%5E%7Bn-k%7D%5C%5C%5C%5C%5C%5CP%28y%3Dk%29%20%3D%20%5Cdbinom%7B25%7D%7Bk%7D%200.7%5E%7Bk%7D%200.3%5E%7B25-k%7D%5C%5C%5C%5C%5C%5C)
a) Then, for Y=10, the probability is:
![P(y=10) = \dbinom{25}{10} p^{10}(1-p)^{15}=3268760*0.0282475249*0.0000000143\\\\\\P(y=10)=0.0013\\\\\\](https://tex.z-dn.net/?f=P%28y%3D10%29%20%3D%20%5Cdbinom%7B25%7D%7B10%7D%20p%5E%7B10%7D%281-p%29%5E%7B15%7D%3D3268760%2A0.0282475249%2A0.0000000143%5C%5C%5C%5C%5C%5CP%28y%3D10%29%3D0.0013%5C%5C%5C%5C%5C%5C)
b) We have to calculate the probability P(Y≤5)
![P(y\leq5)=P(Y=0)+P(Y=1)+...+P(Y=5)\\\\\\P(x=0) = \dbinom{25}{0} p^{0}(1-p)^{25}=1*1*0=0\\\\\\P(y=1) = \dbinom{25}{1} p^{1}(1-p)^{24}=25*0.7*0=0\\\\\\P(y=2) = \dbinom{25}{2} p^{2}(1-p)^{23}=300*0.49*0=0.0000000001\\\\\\P(y=3) = \dbinom{25}{3} p^{3}(1-p)^{22}=2300*0.343*0=0.0000000025\\\\\\P(y=4) = \dbinom{25}{4} p^{4}(1-p)^{21}=12650*0.2401*0=0.0000000318\\\\\\P(y=5) = \dbinom{25}{5} p^{5}(1-p)^{20}=53130*0.16807*0=0.0000003114\\\\\\\\](https://tex.z-dn.net/?f=P%28y%5Cleq5%29%3DP%28Y%3D0%29%2BP%28Y%3D1%29%2B...%2BP%28Y%3D5%29%5C%5C%5C%5C%5C%5CP%28x%3D0%29%20%3D%20%5Cdbinom%7B25%7D%7B0%7D%20p%5E%7B0%7D%281-p%29%5E%7B25%7D%3D1%2A1%2A0%3D0%5C%5C%5C%5C%5C%5CP%28y%3D1%29%20%3D%20%5Cdbinom%7B25%7D%7B1%7D%20p%5E%7B1%7D%281-p%29%5E%7B24%7D%3D25%2A0.7%2A0%3D0%5C%5C%5C%5C%5C%5CP%28y%3D2%29%20%3D%20%5Cdbinom%7B25%7D%7B2%7D%20p%5E%7B2%7D%281-p%29%5E%7B23%7D%3D300%2A0.49%2A0%3D0.0000000001%5C%5C%5C%5C%5C%5CP%28y%3D3%29%20%3D%20%5Cdbinom%7B25%7D%7B3%7D%20p%5E%7B3%7D%281-p%29%5E%7B22%7D%3D2300%2A0.343%2A0%3D0.0000000025%5C%5C%5C%5C%5C%5CP%28y%3D4%29%20%3D%20%5Cdbinom%7B25%7D%7B4%7D%20p%5E%7B4%7D%281-p%29%5E%7B21%7D%3D12650%2A0.2401%2A0%3D0.0000000318%5C%5C%5C%5C%5C%5CP%28y%3D5%29%20%3D%20%5Cdbinom%7B25%7D%7B5%7D%20p%5E%7B5%7D%281-p%29%5E%7B20%7D%3D53130%2A0.16807%2A0%3D0.0000003114%5C%5C%5C%5C%5C%5C%5C%5C)
![P(y\leq5)=0+0+0.0000000001+0.0000000025+0.0000000318+0.00000031\\\\P(y\leq5)= 0.00000035](https://tex.z-dn.net/?f=P%28y%5Cleq5%29%3D0%2B0%2B0.0000000001%2B0.0000000025%2B0.0000000318%2B0.00000031%5C%5C%5C%5CP%28y%5Cleq5%29%3D%200.00000035)
c) The mean and standard deviation for this binomial distribution can be calculated as:
![\mu=np=25\cdot 0.7=17.5\\\\\sigma=\sqrt{np(1-p)}=\sqrt{25\cdot0.7\cdot0.3}=\sqrt{5.25}=2.29](https://tex.z-dn.net/?f=%5Cmu%3Dnp%3D25%5Ccdot%200.7%3D17.5%5C%5C%5C%5C%5Csigma%3D%5Csqrt%7Bnp%281-p%29%7D%3D%5Csqrt%7B25%5Ccdot0.7%5Ccdot0.3%7D%3D%5Csqrt%7B5.25%7D%3D2.29)
Refer to the diagram shown below.
Assume that approximately all the data is contained under the probability distribution curve within 3 standard deviations from the mean. It is actually 99.7%.
Therefore,
bin #1 corresponds to μ-3σ,
bin #11 corresponds to μ,
bin #21 corresponds to μ + 3σ.
Let x be the random variable for bin 17.
Then by interpolation,
![\frac{x-\mu}{(\mu + 3\sigma )-\mu} = \frac{17-11}{21-11} \\\\ x-\mu = 1.8\sigma](https://tex.z-dn.net/?f=%20%5Cfrac%7Bx-%5Cmu%7D%7B%28%5Cmu%20%2B%203%5Csigma%20%29-%5Cmu%7D%20%3D%20%5Cfrac%7B17-11%7D%7B21-11%7D%20%5C%5C%5C%5C%20x-%5Cmu%20%3D%201.8%5Csigma)
The z-score for x is
![z= \frac{x-\mu}{\sigma} = \frac{1.8\sigma}{\sigma} =1.8](https://tex.z-dn.net/?f=z%3D%20%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%20%3D%20%5Cfrac%7B1.8%5Csigma%7D%7B%5Csigma%7D%20%3D1.8)
The probability corresponding to bin #17 is (from standard tables)
P(z≤1.8) = 0.964
Therefore if there are 1000 balls, the sum of the balls in bins 1 to 17 is
0.964 * 1000 = 96.4 ≠ 96 balls.
Answer: 96 balls