Question

for a standard normal distribution, use technology to compute the following probabilities. round answers to 4 decimal places.
a) ( p(z > -0.6069) =)
b) ( p(-2.8394 < z < -0.4507) =)
c) ( p(z < -0.2749) =)
d) ( p(z geq 2.1092) =)
question help: video message instructor

Explanation:

Step1: Use the property of the standard normal distribution

For a standard normal random variable \(Z\), \(P(Z > a)=1 - P(Z\leq a)\), \(P(a < Z < b)=P(Z < b)-P(Z < a)\)

Step2: Calculate each probability using technology (e.g., a TI - 84 Plus calculator or R - software)

• a)

Using the formula \(P(Z > - 0.6069)=1 - P(Z\leq - 0.6069)\)
If we use a TI - 84 Plus: normalcdf(-1E99,-0.6069,0,1) gives \(P(Z\leq - 0.6069)\approx0.2719\)
Then \(P(Z > - 0.6069)=1 - 0.2719 = 0.7281\)

• b)

Using the formula \(P(-2.8394 < Z < - 0.4507)=P(Z < - 0.4507)-P(Z < - 2.8394)\)
normalcdf(-1E99,-0.4507,0,1) gives \(P(Z < - 0.4507)\approx0.3250\)
normalcdf(-1E99,-2.8394,0,1) gives \(P(Z < - 2.8394)\approx0.0023\)
Then \(P(-2.8394 < Z < - 0.4507)=0.3250 - 0.0023=0.3227\)

• c)

Using normalcdf(-1E99,-0.2749,0,1) gives \(P(Z < - 0.2749)\approx0.3918\)

• d)

Using the formula \(P(Z\geq2.1092)=1 - P(Z < 2.1092)\)
normalcdf(-1E99,2.1092,0,1) gives \(P(Z < 2.1092)\approx0.9826\)
Then \(P(Z\geq2.1092)=1 - 0.9826 = 0.0174\)

Answer:

a) \(0.7281\)
b) \(0.3227\)
c) \(0.3918\)
d) \(0.0174\)