Chapter 14, Problem 37SGA

To determine

To find the range about the mean such that the probability is $0.9$ .

Answer to Problem 37SGA

  $\left[0.975,2.625\right]$

Explanation of Solution

Given:

The average hours of cleaning the home is $1.8$ with a standard deviation of $0.5$ in a random sample of $200$ adults.

Calculation:

Given that -

  $\sigma =0.5,N=200,\overline{X}=1.8$

It is also given that $P=90\%$ .

Now, $P=90\%$ correspond to $t=1.65$ using page number 919.

Now, finally let us find the range about the mean as follows-

  $\begin{array}{l}\left[\overline{X}-t\sigma ,\overline{X}+t\sigma \right]\\ =\left[1.8-1.65\times 0.5,1.8+1.65\times 0.5\right]\\ =\left[0.975,2.625\right]\end{array}$