Normal Distribution Worksheet Find the highest and lowest interval that contains about 95% of the distribution. Assume the variables are continuous, random and normally distributed. m = 15, s = 3 m = 19, s = 2.5 m = 4.2, s = .25 m = 15, s = .085 Application of z-scores A brand of automobile tire has a life expectancy that is normally distributed, with a mean life of 30,000 miles and a standard deviation of 2,500 miles. The lifespan of three randomly selected tires are 33,000 miles, 24,750 miles, and 30,000 miles. Find the z-score that corresponds to each lifespan. Your company manufactures ball bearings. The diameters of the ball bearings are normally distributed, with a mean of 3 inches and a standard deviation of 0.02 inch. Find the z-scores of four randomly selected ball bearings with the diameters 3.01, 2.97, 2.98, and 3.05 inches. If you fill an order for 50,000 ball bearings and are willing to replace any whose diameter varies from 3 inches by more than 0.025 inch, how many ball bearings should you expect to replace? The lifespan of a species of fruit fly is normally distributed with a mean of 36 days and a standard deviation of 4 days. The lifespan of three randomly selected flies are 32 days, 35 days, 36 days, and 42 days. Find the z-score that corresponds to each lifespan.

MATLAB: An Introduction with Applications

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Normal Distribution Worksheet

Find the highest and lowest interval that contains about 95% of the distribution. Assume the variables are continuous, random and normally distributed.

m = 15, s = 3
m = 19, s = 2.5
m = 4.2, s = .25
m = 15, s = .085

Application of z-scores

A brand of automobile tire has a life expectancy that is normally distributed, with a mean life of 30,000 miles and a standard deviation of 2,500 miles. The lifespan of three randomly selected tires are 33,000 miles, 24,750 miles, and 30,000 miles. Find the z-score that corresponds to each lifespan.
Your company manufactures ball bearings. The diameters of the ball bearings are normally distributed, with a mean of 3 inches and a standard deviation of 0.02 inch.
1. Find the z-scores of four randomly selected ball bearings with the diameters 3.01, 2.97, 2.98, and 3.05 inches.
2. If you fill an order for 50,000 ball bearings and are willing to replace any whose diameter varies from 3 inches by more than 0.025 inch, how many ball bearings should you expect to replace?
The lifespan of a species of fruit fly is normally distributed with a mean of 36 days and a standard deviation of 4 days. The lifespan of three randomly selected flies are 32 days, 35 days, 36 days, and 42 days. Find the z-score that corresponds to each lifespan.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Expert Solution

Step 1

The given values

1. $μ = 15, σ = 3$

When 95% level of significance the z value is 1.96.

Therefore, the 95% confidence interval is computed as follows:

$\begin{array}{rcl} C . I & = & μ \pm z σ \\ = & 15 \pm 1.96 \times 3 \\ = & (9.12, 20.88) \end{array}$

2. $μ = 19, σ = 2.5$

$\begin{array}{rcl} C . I & = & μ \pm z σ \\ = & 19 \pm 1.96 \times 2.5 \\ = & (14.1, 23.9) \end{array}$

3. $μ = 4.2, σ = 0.25$

$\begin{array}{rcl} C . I & = & μ \pm z σ \\ = & 4.2 \pm 1.96 \times 0.25 \\ = & (3.71, 4.69) \end{array}$

4. $μ = 15, σ = 0.085$

$\begin{array}{rcl} C . I & = & μ \pm z σ \\ = & 15 \pm 1.96 \times 0.085 \\ = & (14.833, 15.167) \end{array}$

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