Respuesta :
Answer:
a) P(50 < x < 54) = 0.112
b) P(55 < x < 62) = 0.428
c) P(53 < x < 70) = 0.644
Step-by-step explanation:
Sample mean = nP = (sample size) × (population proportion) = 150 × 0.4 = 60
Sample standard deviation = √[nP(1-P)] = √(150×0.4×0.6) = 6
To check if this distribution approximates a normal distribution
np = 60 > 10
np(1-p) = 36 > 10
So, this is a normal distribution problem
a) P(50 < x < 54)
We standardize 50 and 54.
The standardized score for any value is the value minus the mean then divided by the standard deviation.
For 50
z = (x - μ)/σ = (50 - 60)/6 = - 1.67
For 54
z = (x - μ)/σ = (54 - 60)/6 = - 1.00
P(50 < x < 54) = P(-1.67 < z < -1.00) = P(z < -1.00) - P(z < -1.67) = 0.159 - 0.047 = 0.11
b) P(55 < x < 62)
We standardize 55 and 62.
The standardized score for any value is the value minus the mean then divided by the standard deviation.
For 55
z = (x - μ)/σ = (55 - 60)/6 = - 0.83
For 62
z = (x - μ)/σ = (54 - 60)/6 = 0.33
P(55 < x < 62) = P(-0.83 < z < 0.33) = P(z < 0.33) - P(z < -0.83) = 0.630 - 0.202 = 0.428
c) P(53 < x < 70)
P(53 < x < 70)
We standardize 53 and 70.
The standardized score for any value is the value minus the mean then divided by the standard deviation.
For 53
z = (x - μ)/σ = (53 - 60)/6 = - 0.50
For 70
z = (x - μ)/σ = (70 - 60)/6 = 1.67
P(53 < x < 70) = P(-0.50 < z < 1.67) = P(z < 1.67) - P(z < -0.50) = 0.953 - 0.309 = 0.644
Hope this Helps!!!
Using the normal approximation to the binomial, it is found that there is:
a) 0.1387 = 13.87% probability of observing between 50 and 54 successes.
b) 0.484 = 48.4% probability of observing between 55 and 62 successes.
c) 0.8543 = 85.43% probability of observing between 53 and 70 successes.
Binomial probability distribution:
Probability of exactly x successes on n trials, with p probability.
Normal Probability Distribution
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 can be approximated to the normal with mean [tex]\mu = np[/tex] and standard deviation [tex]\sigma = \sqrt{np(1-p)}[/tex].
In this problem:
- Sample of 150, thus [tex]n = 150[/tex]
- Proportion of 0.4, thus [tex]p = 0.4[/tex].
For the approximation:
[tex]\mu = np = 150(0.4) = 60[/tex]
[tex]\sigma = \sqrt{np(1-p)} = \sqrt{150(0.4)(0.6)} = 6[/tex]
Item a:
Using continuity correction, this probability is [tex]P(50 - 0.5 \leq X \leq 54 + 0.5) = P(49.5 \leq X \leq 54.5)[/tex], which is the p-value of Z when X = 54.5 subtracted by the p-value of Z when X = 49.5. Thus:
X = 54.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{54.5 - 60}{6}[/tex]
[tex]Z = -0.92[/tex]
[tex]Z = -0.92[/tex] has a p-value of 0.1788
X = 49.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{49.5 - 60}{6}[/tex]
[tex]Z = -1.75[/tex]
[tex]Z = -1.75[/tex] has a p-value of 0.0401.
0.1788 - 0.0401 = 0.1387
0.1387 = 13.87% probability of observing between 50 and 54 successes.
Item b:
This probability is [tex]P(55 - 0.5 \leq X \leq 62 + 0.5) = P(54.5 \leq X \leq 62.5)[/tex], which is the p-value of Z when X = 62.5 subtracted by the p-value of Z when X = 54.5. Thus:
X = 62.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{62.5 - 60}{6}[/tex]
[tex]Z = 0.42[/tex]
[tex]Z = 0.42[/tex] has a p-value of 0.6628.
X = 54.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{54.5 - 60}{6}[/tex]
[tex]Z = -0.92[/tex]
[tex]Z = -0.92[/tex] has a p-value of 0.1788
0.6628 - 0.1788 = 0.484
0.484 = 48.4% probability of observing between 55 and 62 successes.
Item c:
This probability is [tex]P(53 - 0.5 \leq X \leq 70 + 0.5) = P(52.5 \leq X \leq 70.5)[/tex], which is the p-value of Z when X = 70.5 subtracted by the p-value of Z when X = 52.5. Thus:
X = 70.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{70.5 - 60}{6}[/tex]
[tex]Z = 1.75[/tex]
[tex]Z = 1.75[/tex] has a p-value of 0.9599.
X = 52.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{52.5 - 60}{6}[/tex]
[tex]Z = -1.25[/tex]
[tex]Z = -1.25[/tex] has a p-value of 0.1056.
0.9599 - 0.1056 = 0.8543.
0.8543 = 85.43% probability of observing between 53 and 70 successes.
A similar problem is given at https://brainly.com/question/20732994
