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Answer:
Probability that the mean hours spent per week on household chores by a sample of 49 adults will be more than 26.75 is 0.89435.
Step-by-step explanation:
We are given that the number of hours spent per week on household chores by all adults has a mean of 28 hours and a standard deviation of 7 hours.
Also, sample of 49 adults is given.
Let X = number of hours spent per week on household chores
So, assuming data follows normal distribution; X ~ N([tex]\mu=28,\sigma^{2}=7^{2}[/tex])
The z score probability distribution for sample mean is given by;
Z = [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean hours = 28
[tex]\sigma[/tex] = standard deviation = 7 hours
n = sample of adults = 49
Let [tex]\bar X[/tex] = sample mean hours spent per week on household chores
So, probability that the mean hours spent per week on household chores by a sample of 49 adults will be more than 26.75 is given by =P([tex]\bar X[/tex] > 26.75 hours)
P([tex]\bar X[/tex] > 26.75 hours) = P( [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{26.75-28}{\frac{7}{\sqrt{49} } }[/tex] ) = P(Z > -1.25) = P(Z < 1.25)
= 0.89435 {using z table}
Therefore, probability that the mean hours spent per week on household chores is more than 26.75 is 0.89435.
Using the normal distribution and the central limit theorem, it is found that there is a 0.8943 = 89.43% probability that the mean hours spent per week on household chores by a sample of 49 adults will be more than 26.75.
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.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
In this problem:
- Mean of 28 hours, hence [tex]\mu = 28[/tex]
- Standard deviation of 7 hours, hence [tex]\sigma = 7[/tex]
- Sample of 49 adults, hence [tex]n = 49, s = \frac{7}{\sqrt{49}} = 1[/tex]
The probability that the mean hours spent per week on household chores by a sample of 49 adults will be more than 26.75 is 1 subtracted by the p-value of Z when X = 26.75, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{26.75 - 28}{1}[/tex]
[tex]Z = -1.25[/tex]
[tex]Z = -1.25[/tex] has a p-value of 0.1057.
1 - 0.1057 = 0.8943.
0.8943 = 89.43% probability that the mean hours spent per week on household chores by a sample of 49 adults will be more than 26.75.
A similar problem is given at https://brainly.com/question/24663213
