Question

The daily high temperatures in a vacation resort city are approximately Normal, with a mean temperature of 75

Answer 1

Answer:

At 12.10% of days  fewer lifeguards will be provided.

Step-by-step explanation:

We are given that the daily high temperatures in a vacation resort city are approximately Normal, with a mean temperature of 75  degrees Fahrenheit and a standard deviation of 6 degrees.

Let X = the daily high temperatures in a vacation resort city

So, X ~ Normal($\mu=75,\sigma^{2} =6^{2}$)

The z-score probability distribution for the normal distribution is given by;

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

where, $\mu$ = mean temperature = 75 degrees Fahrenheit

            $\sigma$ = standard deviation = 6 degrees Fahrenheit

Now, the percentage of days at which  fewer lifeguards will be provided is given by = P(X $\leq$ 68°)

  P(X $\leq$ 68°)  = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{68-75}{6}$ ) = P(Z $\leq$ -1.17) = 1 - P(Z < 1.17)

                                                  = 1 - 0.879 = 0.121 or 12.10%

The above probability is calculated by looking at the value of x = 1.17 in the z-table which has an area of 0.8790.

Hence, at 12.10% of days  fewer lifeguards will be provided.