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. If a weather forecast predicts the high temperature will be at most 68 degrees, the city provides fewer lifeguards for the city beaches. On what percentage of days will
fewer lifeguards be provided?
Find the z-table here.
O 4.36%
O 12.10%
O 87.90%
O 95.64%

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([tex]\mu=75,\sigma^{2} =6^{2}[/tex])

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

Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = mean temperature = 75 degrees Fahrenheit

[tex]\sigma[/tex] = standard deviation = 6 degrees Fahrenheit

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

P(X [tex]\leq[/tex] 68°) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{68-75}{6}[/tex] ) = P(Z [tex]\leq[/tex] -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.