stats ch.6

Attendance at large exhibition shows in Denver averages about 8100 people per day, with standard deviation of about 500. Assume that the daily attendance figures follow a normal distribution. (Round your answers to four decimal places.)

(a) What is the probability that the daily attendance will be fewer than 7200 people? .0359 (b) What is the probability that the daily attendance will be more than 8900 people? .0548 (c) What is the probability that the daily attendance will be between 7200 and 8900 people? .9093

What percentage of the area under the normal curve lies as given below?

(a) to the right of μ 50% between μ - 2σ and μ + 2σ 95% to the right of μ + 3σ (Use 2 decimal places.) .15

Let z be a random variable with a standard normal distribution. Find the indicated probability. (Round your answer to four decimal places.) P(−1.76 ≤ z ≤ −1.20) =

.0759 shaded in between -2 and -1

Sketch the area under the standard normal curve over the indicated interval and find the specified area. (Round your answer to four decimal places.) The area between z = 1.30 and z = 2.20 is

.0829

Find z such that 12% of the standard normal curve lies to the left of z. Sketch the area described. Step 1 We want to find the z value with 12% of the area under the standard normal curve to the left of z. In other words, we wish to find z such that the area under the curve to the left of z is equal to

.12

Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability. (Round your answer to four decimal places.) μ = 5.5; σ = 1.4 P(7 ≤ x ≤ 9) =

0.1361

Find z such that 87.9% of the standard normal curve lies to the left of z. (Round your answer to two decimal places.) z =

1.17 -z shaded to the left

Find z such that 3.7% of the standard normal curve lies to the right of z. (Round your answer to two decimal places.) z =

1.79 z shaded to the right

(a) Find the z score corresponding to x=31 Recall that the z score gives the number of standard deviations between the original measurement x and the mean μ of the x distribution. It can be calculated using the following equation. z = x − μ σ We are given μ = 35 and σ = 4. Calculate the z score when x = 31. z = x − μ σ

31 -4 -1

b) Find the z score corresponding to x = 45. Now, repeat the process from part (a) and calculate the z score given μ = 35, σ = 4, and x = 45. z = x − μ σ

45 10 2.5

Sketch the area under the standard normal curve over the indicated interval and find the specified area. (Round your answer to four decimal places.) The area to the left of z = −0.46 is

.3228

Find z such that 94.6% of the standard normal curve lies to the right of z. (Round your answer to two decimal places.) z =

-1.60 -z shaded to the right

Sketch the area under the standard normal curve over the indicated interval and find the specified area. (Round your answer to four decimal places.) The area between z = −1.48 and z = 1.95 is

.905

d) Find the raw score corresponding to z=1 .8 Now, repeat the process from part (c) and calculate x given μ = 35, σ = 4, and z = 1.8. x = zσ + μ

1.8 42.2

Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability. (Round your answer to four decimal places.) μ = 2.4; σ = 0.33 P(x ≥ 2) =

.8869

Let z be a random variable with a standard normal distribution. Find the indicated probability. (Round your answer to four decimal places.) P(−2.19 ≤ z ≤ 0) =

.4857 shaded in between -2.1 and 0

Let z be a random variable with a standard normal distribution. Find the indicated probability. (Round your answer to four decimal places.) P(z ≤ 2.90) =

.9981 Shade the corresponding area under the standard normal curve. -2 shaded to the left

b) Since we are dealing with an area to the left of z, we can use the Standard Normal Distribution Table directly. z 0.04 0.05 0.06 0.07 0.08 0.09 −1.1 0.1271 0.1251 0.1230 0.1210 0.1190 0.1170 We scan the table in search of the exact value 0.12. However, this area is not in our table, so we will need to use approximation techniques.

The area 0.1190 is slightly smaller than 0.12 and is associated with the z value -1.18 seenKey -1.18 . The area 0.1210 is slightly larger than 0.12 is associated with the z value -1.17 Correct: Your answer is correct. seenKey -1.17 . The average of these two z values is z = -1.175 which, since 0.12 is halfway between 0.1190 and 0.1210, is the approximated z value associated with the area 0.12. -z shaded to the left