Question

1. The personnel manager of Cumberland Pig Iron Company studies the number of work accidents in a month and prepared the following probability distribution. Calculate the mean, variance, and standard deviation of the number of accidents in one month. Number of accidents Probability 0 0.40 1 0.20 2 0.20 3 0.10 4 0.10 2. Bank of Hawaii reports that 10% of its credit card customers will stop pay at some point. The Hilo branch sent out 12 new cards today. to. How many of the new cardholders do you think will stop paying? Which is the standard deviation? b. What is the probability that none of the cardholders stops pay? c. What is the probability that at least one defaults? 3. A study related to the cash register lines at Safeway Supermarket, in the South Strand area, revealed that between 4 and 7 p.m. On weekends there is an average of four customers in the waiting line. Which is the probability that when visiting Safeway during this time you will find the following: to. no customer in line? b. Four customers in the waiting line? c. Four or fewer customers in line? d. Four or more customers waiting? 4. The accounting department of Weston Materials, Inc., manufacturer of carports detachable, indicates that it takes two construction workers an average of 32 hours, with a standard deviation of two hours, to assemble the Red Barn model. Assume that the assembly times have to. Determine the z values of 29 and 34 hours. What percentage of garages Does it require between 32 and 34 hours of assembly? b. What percentage of garages require between 29 and 34 hours of assembly? c. What percentage of garages require 28.7 hours or less of assembly? d. How many hours does it take to assemble 5% of the garages?

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**Problem 1:**

Given probability distribution of number of accidents in a month:

Number of accidents | Probability

--- | ---

0 | 0.40

1 | 0.20

2 | 0.20

3 | 0.10

4 | 0.10

**Mean:**

The mean of the probability distribution is calculated by multiplying each value by its probability and then summing the products.

\text{Mean}=\sum x\cdot P(x)=0\cdot0.40+1\cdot0.20+2\cdot0.20+3\cdot0.10+4\cdot0.10=1.3

**Variance:**

The variance is calculated as the average of the squared differences from the mean.

\text{Variance}=\sum(x-\mu)^2\cdot P(x)=(0-1.3)^2\cdot0.40+(1-1.3)^2\cdot0.20+(2-1.3)^2\cdot0.20+(3-1.3)^2\cdot0.10+(4-1.3)^2\cdot0.10=1.81

**Standard Deviation:**

The standard deviation is the square root of the variance.

\text{Standard Deviation}=\sqrt{\text{Variance}}=\sqrt{1.81}=1.3454

Therefore, the mean number of accidents in one month is 1.3, the variance is 1.81, and the standard deviation is approximately 1.3454.

\boxed{\text{Mean}=1.3}

\boxed{\text{Variance}=1.81}

\boxed{\text{Standard Deviation}\approx1.3454}

Given probability distribution of number of accidents in a month:

Number of accidents | Probability

--- | ---

0 | 0.40

1 | 0.20

2 | 0.20

3 | 0.10

4 | 0.10

**Mean:**

The mean of the probability distribution is calculated by multiplying each value by its probability and then summing the products.

**Variance:**

The variance is calculated as the average of the squared differences from the mean.

**Standard Deviation:**

The standard deviation is the square root of the variance.

Therefore, the mean number of accidents in one month is 1.3, the variance is 1.81, and the standard deviation is approximately 1.3454.

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