Question

1. For a standard normal distribution, given:

P(z < c) = 0.9353

Find c.

2. A population of values has a normal distribution with μ=195.6μ=195.6 and σ=42.9σ=42.9. You intend to draw a random sample of size n=203n=203.

What is the mean of the distribution of sample means?
μ¯x=μx¯=

What is the standard deviation of the distribution of sample means?
(Report answer accurate to 2 decimal places.)
σ¯x=σx¯=

3.A population of values has a normal distribution with μ=5.8μ=5.8 and σ=80σ=80. You intend to draw a random sample of size n=131n=131.

What is the mean of the distribution of sample means?
μ¯x=μx¯=

What is the standard deviation of the distribution of sample means (i.e. the standard error)?
(Report answer accurate to 2 decimal places.)
σ¯x=σx¯=

4. CNNBC recently reported that the mean annual cost of auto insurance is 991 dollars. Assume the standard deviation is 251 dollars. You take a simple random sample of 61 auto insurance policies.

Find the probability that a single randomly selected value is less than 971 dollars.
P(X < 971) =

Find the probability that a sample of size n=61n=61 is randomly selected with a mean less than 971 dollars.
P(X̅ < 971) =

Enter your answers as numbers accurate to 4 decimal places.

5. A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 19 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 19 weeks and that the population standard deviation is 2.8 weeks. Suppose you would like to select a random sample of 67 unemployed individuals for a follow-up study.

Find the probability that a single randomly selected value is greater than 19.1.
P(X > 19.1) =  (Enter your answers as numbers accurate to 4 decimal places.)

Find the probability that a sample of size n=67n=67 is randomly selected with a mean greater than 19.1.
P(X̅ > 19.1) =  (Enter your answers as numbers accurate to 4 decimal places.)

Answer 1

1. P(z < c) = 0.9353  

Z score at p = 0.9353 using excel = NORM.S.INV(0.9353) = 1.5165

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2. µ = 195.6, σ = 42.9, n = 203

µₓ = 195.6

σₓ = σ/√n = 42.9/√203 = 3.01

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3. µ = 5.8, σ = 80, n = 131

µₓ = 5.8

σₓ = σ/√n = 80/√131 = 6.99

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4. µ = 991, σ = 251, n = 61

a) P(X < 971) =

= P( (X-µ)/σ < (971-991)/251 )

= P(z < -0.0797)

Using excel function:

= NORM.S.DIST(-0.0797, 1)

= 0.4682

b) P(X̅ < 971) =

= P( (X̅-μ)/(σ/√n) < (971-991)/(251/√61) )

= P(z < -0.6223)

Using excel function:

= NORM.S.DIST(-0.6223, 1)

= 0.2669

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5. µ = 19, σ = 2.8, n = 67

a) P(X > 19.1) =

= P( (X-µ)/σ > (19.1-19)/2.8)

= P(z > 0.0357)

= 1 - P(z < 0.0357)

Using excel function:

= 1 - NORM.S.DIST(0.0357, 1)

= 0.4858

b) P(X̅ > 19.1) =

= P( (X̅-μ)/(σ/√n) > (19.1-19)/(2.8/√67) )

= P(z > 0.2923)

= 1 - P(z < 0.2923)

Using excel function:

= 1 - NORM.S.DIST(0.2923, 1)

= 0.385