How do you write 5/20 as a percent?????

The distribution of lifetimes of a particular brand of car tires has a mean of 51,200 miles and a standard deviation of 8,200 mi

Orlov [11]

Answer:

a) 0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

b) 0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

c) 0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

d) 0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

Step-by-step explanation:

Problems of normally distributed distributions are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 51200, \sigma = 8200$

Probabilities:

A) Between 55,000 and 65,000 miles

This is the pvalue of Z when X = 65000 subtracted by the pvalue of Z when X = 55000. So

X = 65000

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{65000 - 51200}{8200}$

$Z = 1.68$

$Z = 1.68$ has a pvalue of 0.954

X = 55000

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{55000 - 51200}{8200}$

$Z = 0.46$

$Z = 0.46$ has a pvalue of 0.677

0.954 - 0.677 = 0.277

0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

B) Less than 48,000 miles

This is the pvalue of Z when X = 48000. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{48000 - 51200}{8200}$

$Z = -0.39$

$Z = -0.39$ has a pvalue of 0.348

0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

C) At least 41,000 miles

This is 1 subtracted by the pvalue of Z when X = 41,000. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{41000 - 51200}{8200}$

$Z = -1.24$

$Z = -1.24$ has a pvalue of 0.108

1 - 0.108 = 0.892

0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

D) A lifetime that is within 10,000 miles of the mean

This is the pvalue of Z when X = 51200 + 10000 = 61200 subtracted by the pvalue of Z when X = 51200 - 10000 = 412000. So

X = 61200

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{61200 - 51200}{8200}$

$Z = 1.22$

$Z = 1.22$ has a pvalue of 0.889

X = 41200

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{41200 - 51200}{8200}$

$Z = -1.22$

$Z = -1.22$ has a pvalue of 0.111

0.889 - 0.111 = 0.778

0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

4 0

3 years ago

A box contains 20 light box of which five or defective it for lightbulbs or pick from the box randomly what's the probability th

Snowcat [4.5K]

Answer:

1

Step-by-step explanation:

Given:-

- The box has n = 20 light-bulbs

- The number of defective bulbs, d = 5

Find:-

what's the probability that at most two of them are defective

Solution:-

- We will pick 2 bulbs randomly from the box. We need to find the probability that at-most 2 bulbs are defective.

- We will define random variable X : The number of defective bulbs picked.

Such that, P ( X ≤ 2 ) is required!

- We are to make a choice " selection " of no defective light bulb is picked from the 2 bulbs pulled out of the box.

- The number of ways we choose 2 bulbs such that none of them is defective, out of 20 available choose the one that are not defective i.e n = 20 - 5 = 15 and from these pick r = 2:

X = 0 , Number of choices = 15 C r = 15C2 = 105 ways

- The probability of selecting 2 non-defective bulbs:

P ( X = 0 ) = number of choices with no defective / Total choices

= 105 / 20C2 = 105 / 190

= 0.5526

- The number of ways we choose 2 bulbs such that one of them is defective, out of 20 available choose the one that are not defective i.e n = 20 - 5 = 15 and from these pick r = 1 and out of defective n = 5 choose r = 1 defective bulb:

X = 1 , Number of choices = 15 C 1 * 5 C 1 = 15*5 = 75 ways

- The probability of selecting 1 defective bulbs:

P ( X = 1 ) = number of choices with 1 defective / Total choices

= 75 / 20C2 = 75 / 190

= 0.3947

- The number of ways we choose 2 bulbs such that both of them are defective, out of 5 available defective bulbs choose r = 2 defective.

X = 2 , Number of choices = 5 C 2 = 10 ways

- The probability of selecting 2 defective bulbs:

P ( X = 2 ) = number of choices with 2 defective / Total choices

= 10 / 20C2 = 10 / 190

= 0.05263

- Hence,

P ( X ≤ 2 ) = P ( X =0 ) + P ( X = 1 ) + P (X =2)

= 0.5526 + 0.3947 + 0.05263

= 1

7 0

3 years ago

Please help me I’ll will be marking brainliest please

lesya [120]

Answer:

it would cost $34.50 to purchase 7.5 lbs of walnuts.

3 0

3 years ago

Read 2 more answers

Can you please assist

ELEN [110]

A. (Number of sides - 2) x 100, so
6-2= 4 x 180
= 720° is the sum of all angles in a hexagon
84 + 87 + 128 + 150 + 103 = 552
720 - 552 = 168
X = 168°

2. Angles in a triangle sum up to 180°
36 + 57= 93
180 - 93 = 87
A= 87°

8 0

2 years ago

(-3x)+40=38 is the answer 3/2 , 2/3 ,-2/3 ,or -3/2

Studentka2010 [4]

I hope this helps you

5 0

3 years ago

Read 2 more answers

How do you write 5/20 as a percent?????​

How do you write 5/20 as a percent?????