what is p(0.6 ≤ z ≤ 2.12)?\n16%\n26%\n73%\n98%\ndone
what is p(0.6 ≤ z ≤ 2.12)?\n16%\n26%\n73%\n98%\ndone
Answer
Answer:
16%
Explanation:
Step1: Recall z - table property
$P(0.6\leq z\leq2.12)=P(z\leq2.12)-P(z\leq0.6)$
Step2: Look up values in z - table
From the standard normal distribution table, $P(z\leq2.12) = 0.9830$ and $P(z\leq0.6)=0.7257$
Step3: Calculate the probability
$P(0.6\leq z\leq2.12)=0.9830 - 0.7257=0.2573\approx 26%$ (There may be a rounding - off difference in the z - table values used. If we use more precise values or a calculator with normalcdf function: normalcdf(0.6, 2.12, 0, 1)=0.2574. Rounding to the nearest 10%, the closest answer is 26%)