Question

Find z such that 87.9% of the standard normal curve lies to the left of z. (round your answer to two decimal places.) z =

Answer 1

From standard normal distribution tables and with area to the left of Z being 87.9% or 0.879,

Z= 1.17 (that is, 1.1 at the left column + 0.07 at the top row where 0.879 is found)

Answer 2

Answer with explanation:

It is given that, 87.9% of the standard normal curve lies to the left of z.

  Z (87.9%)

          $=\frac{87.9}{100}\\\\=0.879$

Total Area under Z curve =1

Now, we have to find area left of, (Z value =87.9%=88%)=0.88

 $Z_{0.88}=0.8106$

⇒Area to left of Z curve , when, (Z value =87.9%=88%)=0.88

     = 0.81 (Approximation to two decimal Places)