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:

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()

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

Let = 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( > 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.

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !