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PROBABILTY AND DISTRIBUTIONS
A elevator manufacturing company believes that 'X' is the amount of that can elevator withstand without any damage with is mean 100 and standard deviation 10. This elevator is used to lift the company staff persons with mean 5 and standard deviation 0.5. How many staff person would have to be in the elevator for the probability of No damage exceeds to 0.85.
asked
Jun 30
in
Mathematical Logic
by
Stanfordboi
(
6
points)

7
views
discrete_maths
probability
poissondistribution
binomialdistribution
0
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0
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GATE2005IT32 Video Solution
An unbiased coin is tossed repeatedly until the outcome of two successive tosses is the same. Assuming that the trials are independent, the expected number of tosses is $3$ $4$ $5$ $6$
asked
Apr 19
in
Probability
by
admin
(
3.6k
points)

2
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gate2005it
probability
binomialdistribution
expectation
normal
videosolution
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0
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GATE2006IT22 Video Solution
When a coin is tossed, the probability of getting a Head is $p, 0 < p < 1$. Let $N$ be the random variable denoting the number of tosses till the first Head appears, including the toss where the Head appears. Assuming that successive tosses are independent, the expected value of $N$ ... $\dfrac{1}{(1  p)}$ $\dfrac{1}{p^{2}}$ $\dfrac{1}{(1  p^{2})}$
asked
Apr 19
in
Probability
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admin
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3.6k
points)

2
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gate2006it
probability
binomialdistribution
expectation
normal
videosolution
0
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0
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GATE200621 Video Solution
For each element in a set of size $2n$, an unbiased coin is tossed. The $2n$ coin tosses are independent. An element is chosen if the corresponding coin toss was a head. The probability that exactly $n$ elements are chosen is $\frac{^{2n}\mathrm{C}_n}{4^n}$ $\frac{^{2n}\mathrm{C}_n}{2^n}$ $\frac{1}{^{2n}\mathrm{C}_n}$ $\frac{1}{2}$
asked
Apr 19
in
Probability
by
admin
(
3.6k
points)

1
view
gate2006
probability
binomialdistribution
normal
videosolution
0
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0
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GATE200552 Video Solution
A random bit string of length n is constructed by tossing a fair coin n times and setting a bit to 0 or 1 depending on outcomes head and tail, respectively. The probability that two such randomly generated strings are not identical is: $\frac{1}{2^n}$ $1  \frac{1}{n}$ $\frac{1}{n!}$ $1  \frac{1}{2^n}$
asked
Apr 19
in
Probability
by
admin
(
3.6k
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0
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gate2005
probability
binomialdistribution
easy
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