Respuesta :
Answer:
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
(c) exactly 30 times?
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
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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 [tex]\mu = np, \sigma = \sqrt{np(1-p)}[/tex].
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 [tex]p = \frac{1}{6}[/tex].
- The pair of dice are rolled 180 times, thus [tex]n = 180[/tex].
The mean and standard deviation are given by:
[tex]\mu = np = 180\frac{1}{6} = 30[/tex]
[tex]\sigma = \sqrt{np(1-p)} = \sqrt{180\frac{1}{6}\frac{5}{6}} = 5[/tex]
Item a:
Using continuity correction, this probability is [tex]P(X \geq 25 - 0.5) = P(X \geq 24.5)[/tex], which is 1 subtracted by the p-value of Z when X = 24.5. Hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{24.5 - 30}{5}[/tex]
[tex]Z = -1.1[/tex]
[tex]Z = -1.1[/tex] 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 [tex]P(33 - 0.5 \leq X \leq 41 + 0.5) = P(32.5 \leq X \leq 41.5)[/tex], 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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{41.5 - 30}{5}[/tex]
[tex]Z = 2.3[/tex]
[tex]Z = 2.3[/tex] has a p-value of 0.9893.
X = 32.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{32.5 - 30}{5}[/tex]
[tex]Z = 0.5[/tex]
[tex]Z = 0.5[/tex] 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 [tex]P(30 - 0.5 \leq X \leq 30 + 0.5) = P(29.5 \leq X \leq 30.5)[/tex], 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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{30.5 - 30}{5}[/tex]
[tex]Z = 0.1[/tex]
[tex]Z = 0.1[/tex] has a p-value of 0.5398.
X = 29.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{29.5 - 30}{5}[/tex]
[tex]Z = -0.1[/tex]
[tex]Z = -0.1[/tex] 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 https://brainly.com/question/19054425
