Question

re than 52 hours? the sample average is less than 47...

According to a study, the average number of hours of TV viewing per household per week in the Philippines is 50.4 hours and the standard deviation is 11.8 hours. If a random sample of 42 households is taken, what is the probability that-

• the sample average is more than 52 hours?
• the sample average is less than 47.5 hours?
• the sample average is less than 40 hours?

The mean of the number of hours of TV viewing per household per week in the Philippines is,

$\mu =$ 50.4 hours.1

The standard deviation is , hours.

The sample given is 6.

The new standard deviation is,

$\begin{array}{rcl}\overline{\sigma }& =& \frac{11.8}{\sqrt{42}}\\ & =& 1.820779.\end{array}$

a. The probability that the sample average is more than 52 hours is,

$\begin{array}{rcl}P\left(\overline{X}>52\right)& =& P\left(Z>\frac{X-\mu }{\sigma }\right)\\ & =& P\left(Z>\frac{52-50.4}{\left(\frac{11.8}{42}\right)}\right)\\ & =& P\left(Z>\frac{1.6}{1.820779}\right)\\ & =& P\left(Z>0.88\right)\\ & =& 0.1894.\end{array}$

b. The probability that the sample average is less than 47.5 hours is,

$\begin{array}{rcl}P\left(\overline{X}<47.5\right)& =& P\left(Z<\frac{X-\mu }{\sigma }\right)\\ & =& P\left(Z<\frac{47.5-50.4}{\left(\frac{11.8}{42}\right)}\right)\\ & =& P\left(Z<-1.59\right)\\ & =& 0.0559.\end{array}$

c. The probability that the sample average is less than 40 hours is,

$\begin{array}{rcl}P\left(\overline{X}<40\right)& =& P\left(Z<\frac{X-\mu }{\sigma }\right)\\ & =& P\left(Z<\frac{40-50.4}{\left(\frac{11.8}{42}\right)}\right)\\ & =& P\left(Z<-5.71\right)\\ & =& 0.\end{array}$

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