Can someone help, please
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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.
<em>Let X = number of hours spent per week on household chores</em>
So, assuming data follows normal distribution; X ~ N()
The z score probability distribution for sample mean is given by;
Z = ~ N(0,1)
where, = mean hours = 28
= standard deviation = 7 hours
n = sample of adults = 49
<em>Let </em><em> = sample mean hours spent per week on household chores</em>
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( > 26.75 hours)
P( > 26.75 hours) = P(
>
) = 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.