1. Hair length is uniformly distributed between 3 cm and 26 cm. If a person is r

Question

1. Hair length is uniformly distributed between 3 cm and 26 cm. If a person is randomly selected, what is the probability that their hair will be less than 5 cm long?

2. The mean weight of jeans in a load of wash is 24 pounds with a standard deviation of 3 lbs. The distribution is normal. a. Use the Empirical Rule to determine the probability that a load of jeans will weigh between 18 and 30 pounds. b. What is the probability that a load of jeans will weigh less than 24 pounds?

3. Determine z-scores for the following and label these in their appropriate location on the standard normal curve: mean, Q3, z=-2, P99

4. A nutritionist believes the average number of fat grams in one serving of ice cream is 14 grams, with a standard deviation of 3.4 grams. If fat grams in ice cream are normally distributed and a serving is randomly chosen, what is the probability that its fat content will be less than 7 grams? Is a 7g serving considered unusual?

5. The average time that lifeguards spend in the sun is 36 hours with a standard deviation of 13 hours. Suppose 30 lifeguards are randomly sampled. Use the Central Limit Theorem to answer the following: a. What is the distribution of the sample means regardless of the original distribution of each sample? b. What is the mean of the sample means? c. What is the probability that the sample mean will be less than 40 hours?

6. Find the 20th percentile for an aerobics workout that is normally distributed and lasts an average of 60 minutes with a standard deviation of 8 minutes. Explain what this number means.

could you please show me steps?

thanks

Explanation / Answer

Ans:

1)Uniform distribution:

P(X<=x)=(x-a)/(b-a)

P(X<=5)=(5-3)/(26-3)=2/23=0.0870

2)mean=24,

standard deviation=3

a)18 and 30 are 2 standard devoations away from the mean.

(24-2*3=24-6=18 and 24+2*3=30)

According to empirical rule,there is 95% probability that a load of jeans will weigh between 18 and 30 pounds.

b)z=(24-24)/3=0

P(z<0)=0.5

3)for mean,z score=0

For ,Q3,it means P(Z<=z)=0.75

z=0.6745

For,P99 means,P(Z<=z)=0.99

z=2.33

4)mean=14,

standard deviation=3.4

z=(7-14)/3.4=-2.059

P(z<=-2.059)=0.0197

As,0.0197<0.05,so yes it is unusual.

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