Question

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. 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:

Answer 1

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($\mu=28,\sigma^{2}=7^{2}$)

The z score probability distribution for sample mean is given by;

               Z = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = mean hours = 28

            $\sigma$ = standard deviation = 7 hours

            n = sample of adults = 49

Let $\bar X$ = 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($\bar X$ > 26.75 hours)

    P($\bar X$ > 26.75 hours) = P( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{26.75-28}{\frac{7}{\sqrt{49} } }$ ) = 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.

Answer 2

Answer:

0.89435

Step-by-step explanation: