Question

I'm not sure on how to get P(Z < 1.555) from 6%Suppose the time to complete a 200-meter backstroke swim for female competitive swimmers is normally distributed with a mean μ = 141 seconds and a standard deviation σ = 7 seconds. Suppose the fastest 6% of female swimmers in the nation are offered college scholarships. In order to be given a scholarship, a swimmer must complete the 200-meter backstroke in no more than how many seconds? Give your answer in whole numbers.

Answer 1

Answer:

$130\text{ seconds}$

Explanation:

Here, we want to get the time a swimmer must complete the backstroke, in order to be considered for a scholarship

What we need to find here is the time in which the probability would be less than 6% (or 6/100 = 0.06)

Now, let us get the z-score to the right of 0.06 but closest to it

By using the normal distribution table, we have this value as:

$1.55\text{ SD below the mean}$

Recall, mathematically:

$z-score\text{ = }\frac{x-\operatorname{mean}}{SD}$

Substituting the values, we have it that:

$\begin{gathered} -1.55\text{ = }\frac{x-141}{7} \\ x\text{ = 141-7(1.55)} \\ x\text{ = }130.15\text{ } \end{gathered}$

The value of x can be approximated as 130 seconds