Mathematics, 12.11.2020 19:10 isabellam646

A pair of dice is rolled 180 times. What is the probability that a total of 7 occurs
(a) at least 25 times?
(b) between 33 and 41 times inclusive?
(c) exactly 30 times?

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Answer from: seonna

Using the normal approximation to the binomial, it is found that there is a:

a) 0.8643 = 86.43% probability that a total of 7 occurs at least 25 times.

b) 0.2978 = 29.78% probability that a total of 7 occurs between 33 and 41 times inclusive.

c) 0.0796 = 7.96% probability that a total of 7 occurs exactly 30 times.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

In this problem:

When 2 dice are rolled, there are 36 possible outcomes, of which 6 of them[(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)] result in a sum of 7, thus the parameter p is $p = \frac{1}{6}$ .The pair of dice are rolled 180 times, thus n = 180

The mean and standard deviation are given by:

$\mu = np = 180\frac{1}{6} = 30$

$\sigma = \sqrt{np(1-p)} = \sqrt{180\frac{1}{6}\frac{5}{6}} = 5$

Item a:

Using continuity correction, this probability is $P(X \geq 25 - 0.5) = P(X \geq 24.5)$ , which is 1 subtracted by the p-value of Z when X = 24.5. Hence:

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

$Z = \frac{24.5 - 30}{5}$

Z = -1.1

Z = -1.1 has a p-value of 0.1357.

1 - 0.1357 = 0.8643.

0.8643 = 86.43% probability that a total of 7 occurs at least 25 times.

Item b:

Using continuity correction, this probability is $P(33 - 0.5 \leq X \leq 41 + 0.5) = P(32.5 \leq X \leq 41.5)$ , which is the p-value of Z when X = 41.5 subtracted by the p-value of Z when X = 32.5. Hence:

X = 41.5

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

$Z = \frac{41.5 - 30}{5}$

Z = 2.3

Z = 2.3 has a p-value of 0.9893.

X = 32.5

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

$Z = \frac{32.5 - 30}{5}$

Z = 0.5

Z = 0.5 has a p-value of 0.6915.

0.9893 - 0.6915 = 0.2978.

0.2978 = 29.78% probability that a total of 7 occurs between 33 and 41 times inclusive.

Item c:

Using continuity correction, this probability is $P(30 - 0.5 \leq X \leq 30 + 0.5) = P(29.5 \leq X \leq 30.5)$ , which is the p-value of Z when X = 30.5 subtracted by the p-value of Z when X = 29.5. Hence:

X = 30.5

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

$Z = \frac{30.5 - 30}{5}$

Z = 0.1

Z = 0.1 has a p-value of 0.5398.

X = 29.5

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

$Z = \frac{29.5 - 30}{5}$

Z = -0.1

Z = -0.1 has a p-value of 0.4602.

0.5398 - 0.4602 = 0.0796.

0.0796 = 7.96% probability that a total of 7 occurs exactly 30 times.

A similar problem is given at link

Answer from: cristinamiu8169

0.86433

0.29782

0.07966

Step-by-step explanation:

Given that :

Number of trials (n) = 180

Probability that a total of 7 occurs

To obtain a total of 7 from a pair of dice :

Combinations:

(1,6) ; (2,5) ; (3,4) ; (4,3) ; (5,2) ;(6, 1)

Sample space when a pair of dice is rolled = 6² = 36

Probability of success (p) = number of 7

combination possible ÷ sample space)

p = 6 / 36 = 0.1667

q = p' = 1 - p = 1 - 0.1667 = 0.8333

Using Normal approximation :

Mean (m) = np = 180 * 0.1667 = 30.006

Standard deviation (σ) = √npq

σ = √(180 * 0.1667 * 0.8333)

σ = 5.0003999

σ = 5.00

(a) at least 25 times?

P(x ≥ 25)

x = 25 - 0.5 = 24.5

Obtain the Zscore :

Z = (x - m) / σ

Z = (24.5 - 30) / 5

Z = -5.5/5

Z = -1.1

P(z ≤ - 1.1)

USing the z probability calculator :

P(z≥ - 1.1) = 0.86433

(b) between 33 and 41 times inclusive?

P(32.5 ≤ x ≤ 41.5)

Z =[ (32.5 - 30) / 5] ≤ x ≤ [ (41.5 - 30) / 5]

Z = 2.5/5 ≤ x ≤ 11.5 /5

Z = 0.5 ≤ x ≤ 2.3

Using z probability calculator :

P(z ≤ 0.5) = 0.69146

P(z ≤ 2.3) = 0.98928

0.98928 - 0.69146 = 0.29782

P(x = 30)

P(29.5 ≤ x ≤ 30.5)

Z =[ (29.5 - 30) / 5] ≤ x ≤ [ (30.5 - 30) / 5]

Z = - 0.5/5 ≤ x ≤ 0.5 /5

Z = - 0.1 ≤ x ≤ 0.1

Using z probability calculator :

P(z ≤ - 0.1) = 0.46017

P(z ≤ 0.1) = 0.53983

0.53983 - 0.46017 = 0.07966

Answer from: Quest

i need the options

step-by-step explanation:

Answer from: Quest

Where’s the problems ? ?

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A pair of dice is rolled 180 times. What is the probability that a total of 7 occurs
(a) at l...

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