Question #349919

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?

1
Expert's answer
2022-06-15T14:01:25-0400

Solution:

1)

μ=5;\mu=5; - mean;

σ=0.4\sigma=0.4 - standard deviation;

Let's find XX at z=1.3;z=1.3;

z=Xμσ;z=\frac{X-\mu}{\sigma};

X=zσ+μ=1.3×0.4+5=5.52;X=z\sigma+\mu=1.3\times0.4+5=5.52; So, z=1.3z=1.3 score 0.4032. It means 40.32% of X values is less than 5.52. And when X<5.52;X<5.52; then z<1.3z<1.3. So, the result:

X0=5.52;X_0=5.52;

2)

μ=241;\mu=241; σ=2;\sigma=2; X0=235;X_0=235;

Let"s find zz :

z=X0μσ=2352412=3;z=\frac{X_0-\mu}{\sigma}=\frac{235-241}{2}=-3; when z=3z=-3 the score equal 0.0013. It means 0.13% of X values is less than 235. So, 10.0013=0.9987;1-0.0013=0.9987;

P(X235)=99.87%.P(X\ge235)=99.87\%.

3)

μ=P338.1/h;\mu=P338.1/h;

σ=P23.41;\sigma=P23.41;

X0=351.1;X_0=351.1;

Let's find z:z:

z=X0μσ=351.1338.123.41=0.55;z=\frac{X_0-\mu}{\sigma}=\frac{351.1-338.1}{23.41}=0.55; when z=0.55z=0.55 the score 0.7088. It means 70.88% of workers' earn less than P351.1. So, 10.7088=0.2912;1-0.7088=0.2912;

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

Answer:

1) X0=5.52;X_0=5.52;

2) P(X235)=99.87%;P(X\ge235)=99.87\%;

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


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