Question

The probability is 51​% that the sample mean will be between what two​ values, symmetrically distributed around the population​ mean? The lower bound is nothing inches. ​(Round to two decimal places as​ needed.) The upper bound is nothing inches. ​(Round to two decimal places as​ needed.)

Answer 1

Answer:

The lower bound is, $-z=-0.69$ and the upper bound is $z=0.69$.

Step-by-step explanation:

Let the random variable X follows a normal distribution with mean μ and standard deviation σ.

The the random variable Z, defined as $Z=\frac{X-\mu}{\sigma}$ is standardized random variable also known as a standard normal random variable. The random variable  $Z\sim N(0, 1)$.

The standard normal random variable has a symmetric distribution.

It is provided that $P(-z\leq Z\leq z)=0.51$.

Determine the upper and lower bound as follows:

$P(-z\leq Z\leq z)=0.51\\P(Z\leq z)-P(Z\leq -z)=0.51\\P(Z\leq z)-[1-P(Z\leq z)]=0.51\\2P(Z\leq z)-1=0.51\\2P(Z\leq z)=1.51\\P(Z\leq z)=0.755$

Use a standard normal table to determine the value of z.

The value of z such that P (Z ≤ z) = 0.755 is 0.69.

The lower bound is, $-z=-0.69$ and the upper bound is $z=0.69$.