Answer to Question #349919 in Statistics and Probability for Ttypre

Question

1. Suppose X is normally distributed with a mean of 5 and a standard deviation of 0.4. Using the standard score formula, we find P (X ≤ X0) = P (Z ≤ 1.3). What is the value of X0?

2. What will be the value of P (X ≥ 235) given that a normal distribution has µ = 241 and d = 2?

3.The average hourly wage of workers at a fast food restaurant is P338.10/hr. Assume the wages are normally distributed with a standard deviation of P23.41. If a worker at this fast food restaurant is selected at random, what is the probability that the worker earns more than P351.10?

Answer 1

Solution:

1)

$\mu=5;$ - mean;

$\sigma=0.4$ - standard deviation;

Let's find $X$ at $z=1.3;$

$z=\frac{X-\mu}{\sigma};$

$X=z\sigma+\mu=1.3\times0.4+5=5.52;$ So, $z=1.3$ score 0.4032. It means 40.32% of X values is less than 5.52. And when $X<5.52;$ then $z<1.3$. So, the result:

$X_0=5.52;$

2)

$\mu=241;$ $\sigma=2;$ $X_0=235;$

Let"s find $z$ :

$z=\frac{X_0-\mu}{\sigma}=\frac{235-241}{2}=-3;$ when $z=-3$ the score equal 0.0013. It means 0.13% of X values is less than 235. So, $1-0.0013=0.9987;$

$P(X\ge235)=99.87\%.$

3)

$\mu=P338.1/h;$

$\sigma=P23.41;$

$X_0=351.1;$

Let's find $z:$

$z=\frac{X_0-\mu}{\sigma}=\frac{351.1-338.1}{23.41}=0.55;$ when $z=0.55$ the score 0.7088. It means 70.88% of workers' earn less than P351.1. So, $1-0.7088=0.2912;$

$P(X>351.1)=29.12\%;$

Answer:

1) $X_0=5.52;$

2) $P(X\ge235)=99.87\%;$

3) $P(X>351.1)=29.12\%.$