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jolli1 [7]
3 years ago
12

Rolling a number less than 4 on a number cube labeled 1 through 6

Mathematics
1 answer:
olga_2 [115]3 years ago
3 0
Rolling a number less than 4 means that rolling a 1, 2, or 3 will satisfy the requirement. Since 3 of 6 possible outcomes will satisfy the requirement then the likelihood that it will be rolled is 3/6 times. 3/6 reduces to 1/2 or 50% chance.
You might be interested in
According to a Yale program on climate change communication survey, 71% of Americans think global warming is happening.† (a) For
SpyIntel [72]

Answer:

a) 0.2741 = 27.41% probability that at least 13 believe global warming is occurring

b) 0.7611 = 76.11% probability that at least 110 believe global warming is occurring

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.71

(a) For a sample of 16 Americans, what is the probability that at least 13 believe global warming is occurring?

Here n = 16, we want P(X \geq 13). So

P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 13) = C_{16,13}.(0.71)^{13}.(0.29)^{3} = 0.1591

P(X = 14) = C_{16,14}.(0.71)^{14}.(0.29)^{2} = 0.0835

P(X = 15) = C_{16,15}.(0.71)^{15}.(0.29)^{1} = 0.0273

P(X = 16) = C_{16,16}.(0.71)^{16}.(0.29)^{0} = 0.0042

P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) = 0.1591 + 0.0835 + 0.0273 + 0.0042 = 0.2741

0.2741 = 27.41% probability that at least 13 believe global warming is occurring

(b) For a sample of 160 Americans, what is the probability that at least 110 believe global warming is occurring?

Now n = 160. So

\mu = E(X) = np = 160*0.71 = 113.6

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{160*0.71*0.29} = 5.74

Using continuity correction, this is P(X \geq 110 - 0.5) = P(X \geq 109.5), which is 1 subtracted by the pvalue of Z when X = 109.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{109.5 - 113.6}{5.74}

Z = -0.71

Z = -0.71 has a pvalue of 0.2389

1 - 0.2389 = 0.7611

0.7611 = 76.11% probability that at least 110 believe global warming is occurring

3 0
2 years ago
You want to find the mean weight of the apples in a barrel. what do you need to do?
Sveta_85 [38]
If you want to find the weight of the apples in the barrel first you have to ( FIND THE TOTAL WEIGHT OF ALL THE APPLES AND DIVIDE BY ALL THE NUMBER OF APPLES)
4 0
3 years ago
Rachel drove 756 miles in 12 hours.<br> At the same rate, how many miles would she drive in 8 hours?
kvv77 [185]

Answer:

504 miles

Step-by-step explanation:

you would do 756÷12 which equals 63.... For every hour that you drive its 63 miles. Then you do 63×8 to get 504

4 0
3 years ago
Your savings account pays 2.5% interest per year. How much interest does the account pay per month.
Mkey [24]

Answer:

0.21%

Step-by-step explanation:

2.5/12=0.208 or 0.21

3 0
3 years ago
At an airport, 76% of recent flights have arrived on time. A sample of 11 flights is studied. Find the probability that no more
I am Lyosha [343]

Answer:

The probability is  P( X \le 4 ) = 0.0054

Step-by-step explanation:

From the question we are told that

   The percentage that are on time is  p =  0.76

   The  sample size is n =  11

   

Generally the percentage that are not on time is

     q =  1- p

     q =  1-  0.76

     q = 0.24

The  probability that no more than 4 of them were on time is mathematically represented as

        P( X \le 4 ) =  P(1 ) +  P(2) + P(3) +  P(4)

=>     P( X \le 4 ) =  \left n } \atop {}} \right.C_1 p^{1}  q^{n- 1} +   \left n } \atop {}} \right.C_2p^{2}  q^{n- 2} +  \left n } \atop {}} \right.C_3 p^{3}  q^{n- 3}  +  \left n } \atop {}} \right.C_4 p^{4}  q^{n- 4}

P( X \le 4 ) =  \left 11 } \atop {}} \right.C_1 p^{1}  q^{11- 1} +   \left 11 } \atop {}} \right.C_2p^{2}  q^{11- 2} +  \left 11 } \atop {}} \right.C_3 p^{3}  q^{11- 3}  +  \left 11 } \atop {}} \right.C_4 p^{4}  q^{11- 4}

P( X \le 4 ) =  \left 11 } \atop {}} \right.C_1 p^{1}  q^{10} +   \left 11 } \atop {}} \right.C_2p^{2}  q^{9} +  \left 11 } \atop {}} \right.C_3 p^{3}  q^{8}  +  \left 11 } \atop {}} \right.C_4 p^{4}  q^{7}

= \frac{11! }{ 10! 1!}  (0.76)^{1}  (0.24)^{10} +   \frac{11!}{9! 2!}  (0.76)^2 (0.24)^{9} + \frac{11!}{8! 3!}  (0.76)^{3}  (0.24)^{8}  + \frac{11!}{7!4!}  (0.76)^{4}  (0.24)^{7}

P( X \le 4 ) = 0.0054

4 0
3 years ago
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