Question

Given a normal distribution with mu equals 100μ=100 and sigma equals 10 commaσ=10, complete parts​ (a) through​ (d). loading... click here to view page 1 of the cumulative standardized normal distribution table. loading... click here to view page 2 of the cumulative standardized normal distribution table.
a. what is the probability that upper x greater than 85x>85​? the probability that upper x greater than 85x>85 is nothing.

Answer 1

I believe the parts are:

A. What is the probability that Upper X greater than 95X>95​?

B. What is the probability that Upper X less than 75X<75?

C.What is the probability that Upper X less than 85X<85 and Upper X greater than 125X>125?

D. 95​% of the values are between what two​ X-values (symmetrically distributed around the​ mean)?

 

Solution:

We use the equation for z score:

z = (X – μ) / σ

Then use the standard normal probability tables to locate for the value of P at indicated z score value.

 

A. Z = (95 – 100) / 10 = - 0.5

Using the tables, the probability at z = -0.5 using right tailed test is:

P = 0.6915

 

 

B. Z = (75 – 100) / 10 = - 2.5

Using the tables, the probability at z = -2.5 using left tailed test is:

P = 0.0062

 

 

C. Z = (85 – 100) / 10 = - 1.5

Using the tables, the probability at z = -2.5 using left tailed test is:

P = 0.0668

 

Z = (125 – 100) / 10 = 2.5

Using the tables, the probability at z = 2.5 using right tailed test is:

P = 0.0062

 

So the probability that 85<X and X>125 is:

P(total) = 0.0668 + 0.0062

P(total) = 0.073

 

 

D. P(left) = 0.025, Z = -1.96

P(right) = 0.975, Z = 1.96

 

The X’s are calculated using the formula:

X = σz + μ

 

At Z = -1.96

X = 10 (-1.96) + 100 = 80.4

 

At Z = 1.96

X = 10 (1.96) + 100 = 119.6

 

So 95% of the values are between 80.4 and 119.6 (80.4< X <119.6).