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.

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 85X125?

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 85125 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).