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3 years ago

The subset of the sample space probability experiment

Mathematics

1 answer:

leva [86]3 years ago

3 0

Step-by-step explanation:

The sample space of a random experiment is the collection of all possible outcomes. An event associated with a random experiment is a subset of the sample space. The probability of any outcome is a number between 0 and 1. The probabilities of all the outcomes add up to 1.

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A woman bought 100 Christmas cards. For the ones that sing a song when you open them, she paid 30 cents each. For the rest she p

lapo4ka [179]

.05c+.3s=10.25, c+s=100, c=100-s (c is for cards that don't sing and s is for those that do :P)

.05c+.3s=10.25 and c=100-s makes the equation become:

.05(100-s)+.3s=10.25

5-.05s+.3s=10.25

.25s=5.25

s=21, and since c=100-s

c=79

So she bought 21 of the more expensive singing cards...

8 0

3 years ago

Read 2 more answers

The amount of lateral expansion (mils) was determined for a sample of n = 7 pulsed-power gas metal arc welds used in LNG ship co

zvonat [6]

Answer:

95% Confidence interval for σ2 and for σ is (3.33 , 38.85) and (1.82 , 6.23) respectively.

Step-by-step explanation:

We are given that the amount of lateral expansion (mils) was determined for a sample of n = 7 pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was s = 2.83 mils.

Assuming data follows normal distribution.

So, firstly the pivotal quantity for 95% confidence interval for the population variance is given by;

P.Q. = $\frac{(n-1)s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

where, s = sample standard deviation = 2.83 mils

$\sigma^{2}$ = population variance

$\sigma$ = population standard deviation

n = sample size = 7

So, 95% confidence interval for population variance, $\sigma^{2}$ is;

P(1.237 < $\chi^{2} __n_-_1$ < 14.45) = 0.95 {As the table of at 6 degree of freedom

gives critical values of 1.237 & 14.45}

P(1.237 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 14.45) = 0.95

P( $\frac{ 1.237}{(n-1)s^{2} }$ < $\frac{1 }{\sigma^{2} }$ < $\frac{ 14.45}{(n-1)s^{2} }$ ) = 0.95

P( $\frac{ (n-1)s^{2}}{14.45 }$ < $\sigma^{2}$ < $\frac{ (n-1)s^{2}}{1.237 }$ ) = 0.95

95% confidence interval for $\sigma^{2}$ = ( $\frac{ (n-1)s^{2}}{14.45 }$ , $\frac{ (n-1)s^{2}}{1.237 }$ )

= ( $\frac{ (7-1) \times 2.83^{2}}{14.45 }$ , $\frac{ (7-1) \times 2.83^{2}}{1.237 }$ )

= (3.33 , 38.85)

95% C.I. for population standard deviation, $\sigma$ = ( $\sqrt{3.33}$ , $\sqrt{38.85}$ )

= (1.82 , 6.23)

Therefore, 95% confidence interval for the population variance (σ2) and population standard deviation (σ) are (3.33 , 38.85) and (1.82 , 6.23) respectively.

7 0

3 years ago

Solve the following equality. 7t+15>3t-13

alukav5142 [94]

7t + 15 > 3t - 13

Subtract 15 over to -13
Subtract 3t over to 7t

4t > -28

t > -7

4 0

4 years ago

Read 2 more answers

Convert 84% to a fraction in simplest form

MrRa [10]

The simplest form is 21/25

3 0