Respuesta :
Answer:
a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.
b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.
c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set 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]
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 p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
The mean of a population is 74 and the standard deviation is 15.
This means that [tex]\mu = 74, \sigma = 15[/tex]
Question a:
Sample of 36 means that [tex]n = 36, s = \frac{15}{\sqrt{36}} = 2.5[/tex]
This probability is 1 subtracted by the pvalue of Z when X = 78. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{78 - 74}{2.5}[/tex]
[tex]Z = 1.6[/tex]
[tex]Z = 1.6[/tex] has a pvalue of 0.9452
1 - 0.9452 = 0.0548
0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.
Question b:
Sample of 150 means that [tex]n = 150, s = \frac{15}{\sqrt{150}} = 1.2247[/tex]
This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So
X = 77
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{77 - 74}{1.2274}[/tex]
[tex]Z = 2.45[/tex]
[tex]Z = 2.45[/tex] has a pvalue of 0.9929
X = 71
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{71 - 74}{1.2274}[/tex]
[tex]Z = -2.45[/tex]
[tex]Z = -2.45[/tex] has a pvalue of 0.0071
0.9929 - 0.0071 = 0.9858
0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.
c. A random sample of size 219 yielding a sample mean of less than 74.2
Sample size of 219 means that [tex]n = 219, s = \frac{15}{\sqrt{219}} = 1.0136[/tex]
This probability is the pvalue of Z when X = 74.2. So
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{74.2 - 74}{1.0136}[/tex]
[tex]Z = 0.2[/tex]
[tex]Z = 0.2[/tex] has a pvalue of 0.5793
0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2
