An office manager has received a report from a consultant that includes a section on equipment replacement. The report indicates that scanners have a service life that is normally distributed with a mean of 41 months and a standard deviation of 4 months. On the basis of this information, determine the percentage of scanners that can be expected to fail in the following time periods:

We solve the above question using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean = 41 months

σ is the population standard deviation = 4 months

a. Before 38 months of service

Before in z score score means less than 38 months

Hence,

z = 38 - 41/4

z = -0.75

Probability value from Z-Table:

P(x<38) = 0.22663

Converting to percentage = 0.22663 × 100

= 22.663%

b. Between 40 and 45 months of service

For x = 40 months

z = 40 - 41/4

z = -0.2

Probaility value from Z-Table:

P(x =40) = 0.40129

For x = 45

z = 45 - 41/4

z = 1

Probability value from Z-Table:

P(x = 45) = 0.84134

Between 40 and 45 months of service

= 0.84134 - 0.40129

= 0.44005

Converting to Percentage

= 0.44005 × 100

= 44.005%

= 44%

c. Wihin ± 2 months of the mean life

+ 2 months = 41 months + 2 months

= 43 month

- 2 months = 41 months - 2 months

= 39 months

For x = 43

z = 43 - 41 /4

z = 0.5

P-value from Z-Table:

P(x = 43) = 0.69146

For x = 39

z = 39 - 41/4

z = -2/4

z = -0.5

Probability value from Z-Table:

P(x = 39) = 0.30854

Within ± 2 months of the mean life

= 0.69146 - 0.30854

= 0.38292

= 38.3%

Learn more about z-score:

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4 0

2 years ago

An investigator compares the durability of two different compounds used in the manufacture of a certain automobile brake lining.

Tanya [424]

Answer:

98% Confidencce Interval is ( 3030.6, 7467.4 )

Step-by-step explanation:

Given that:

Sample size $n_1 =$ 71

Sample size $n_2 =$ 31

Sample mean $\overline x_1 =$ 41628

Sample mean $x_2 =$ 36,379

Population standard deviation $\sigma_1$ = 4934

Population standard deviation $\sigma_2 =$ 4180

At 98% confidence interval level, the level of significcance = 1 - 0.98 = 0.02

Critical value at $z_{0.02/2} = 2.33$

The Margin of Error = $z \times \sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma^2_2}{n_2} }$

= $2.33 \times \sqrt{\dfrac{4934^2}{71}+\dfrac{4180^2}{31} }$

= $2.33 \times \sqrt{\dfrac{24344356}{71}+\dfrac{17472400}{31} }$

= $2.33 \times \sqrt{906504.06 }$

= 2218.40

The Lower limit = $( \overline x_1 - \overline x_2) - (Margin \ of \ error)$

= ( 41628 - 36379 ) - ( 2218.40)

= 5249 - 2218.40

= 3030.6

The upper limit = $( \overline x_1 - \overline x_2) + (Margin \ of \ error)$

= ( 41628 - 36379 ) + ( 2218.40)

= 5249 + 2218.40

= 7467.4

∴ 98% Confidencce Interval is ( 3030.6, 7467.4 )

5 0

3 years ago

Find a.<br><br> Write your answer in simplest radical form.

weqwewe [10]

Answer:95

Nah I'm playing I dont know

7 0

2 years ago

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