Question

1) The amount of time to complete a physical activity in a PE class is approximately normally normally distributed with a mean of 39.7 seconds and a standard deviation of 7.8 seconds.

a) 80% of all students finish the activity in less than ______ seconds.

2) Heights of 10 year old children, regardless of sex, closely follow a normal distribution with mean 55 inches and standard deviation 6.4 inches. Round answers to 4 decimal places.

a) 75% of all 10 year old child are less than ______ inches.

3) Assume that z-scores are normally distributed with a mean of 0 and a standard deviation of 1.

If P(0<z<a)=0.2764P(0<z<a)=0.2764, find a.

a= ______

4) Adult men have heights with a mean of 69.0 inches and a standard deviation of 2.8 inches. Find the height of a man with a z-score of 3.5 (to 2 decimal places)

5) A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 193.4-cm and a standard deviation of 2.4-cm.

Find P₃₉, which is the length separating the shortest 39% rods from the longest 61%.
P₃₉ = ______-cm

Enter your answer as a number accurate to 1 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

6) For a standard normal distribution, given:

P(z < c) = 0.5605

Find c.

7) A dishwasher has a mean lifetime of 12 years with an estimated standard deviation of 1.25 years. Assume the lifetime of a dishwasher is normally distributed.

Round the probabilities to four decimal places.

It is possible with rounding for a probability to be 0.0000.

A) Find the probability that randomly selected dishwasher has a lifetime that is at most 8.875 years.

B) Is a lifetime of 8.875 years unusually low for a randomly selected dishwasher? Why or why not?

C) What lifetime do 74% of all dishwashers have less than? Round your answer to two decimal places in the first box. Put the correct units in the second box.

Answer 1

1)

Z score corresponding to 80th percentile = 0.8418

Thus, 80% of all students finish the activity in less than 39.7 + 0.8418*7.8 = 46.27 seconds

2)

Z score corresponding to 75th percentile = 0.6745

Thus, 75% of all 10 year old child are less than 55 + 0.6755*6.4 = 59.3168 inches

3)

P(0 < Z < a) = 0.2764

-> a = 0.76

4)

The height of a man with a z-score of 3.5 = 69 + 3.5*2.8

= 78.8 inches

5)

Corresponding to P39, the critical Z score = -0.279

Thus, P39 = 193.4 - 0.279*2.4 = 192.73 cm

6)

P(Z < c) = 0.5605

-> c = 0.1522

7)

A)

P(X ≤ 8.875)

= P{Z ≤ (8.875 - 12)/1.25}

= P(Z < -2.5) = 0.0062

B)

Yes, this is unusually low since the probability is less than 0.05

C)

Corresponding Z score for 74th percentile = 0.6433

The required lifetime = 12 + 0.6433*1.25 = 12.804 years