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Kruka [31]
3 years ago
13

An accelerated life test on a large number of type-D alkaline batteries revealed that the mean life for a particular use before

they failed is 19.0 hours. The distribution of the lives approximated a normal distribution. The standard deviation of the distribution was 1.2 hours. About 95.44 percent of the batteries failed between what two values?a. 8.9 and 18.9
b. 12.2 and 14.2
c. 14.1 and 22.1
d. 16.6 and 21.4
Mathematics
1 answer:
joja [24]3 years ago
8 0

Answer:

option (d) 16.6 and 21.4

Step-by-step explanation:

Data provided in the question:

The mean life for a particular use before they failed = 19.0 hours

The distribution of the lives approximated a normal distribution

The standard deviation of the distribution = 1.2 hours

To find:

The values between which 95.44 percent of the batteries failed

Now,

In Normal distribution, the approximately 95% ( ≈ 95.44% of all values ) falls within 2 standard deviations of the mean

Therefore,

Upper limit = Mean + 2 × standard deviation

⇒ Upper limit = 19.0 + 2 × 1.2 = 21.4

Lower  limit = Mean - 2 × standard deviation

⇒ Lower  limit = 19 - 2 × 1.2 = 16.6

Hence,

the answer is option (d) 16.6 and 21.4

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An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" () and "tails" () which we write , , etc. For ea
boyakko [2]

Answer:

Some details are missing

Step-by-step explanation:

An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails) (t) which we write hth, ttt, etc. For each outcome, let R be the random variable counting the number of heads in each outcome. For example, if the outcome is hht, then R(hht) = 2. Suppose that the random variable X is defined in terms of R as follows: X = 2R² - 6R - 1. The values of X are thus:

Outcome: || Value of X

tht || -5

thh || -5

hth || -5

htt || -5

hhh || -1

tth || -5

hht || -5

ttt || -1

Calculate the probability distribution function of X, i.e. the function Px (x). First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row.

Solution

To calculate the probability distribution function of X.

We have to observe the total outcomes to check the number of "Heads (h) " in each outcome.

The first, fourth and, sixth outcome has 1 head (h)

The second, third and seventh outcome has 2 head (hh)

The fifth outcome has 3 head (hhh)

The eight outcome has 0 appearance of h

We then solve the probability of each occurrence

i.e. The probability of having h, hh, hhh and no occurrence of h

This will be represented as follows

P(h=0)

P(h=1)

P(h=2)

P(h=3)

In a coin, the probability of getting a head = ½ and the probability of getting a tail = ½ in 1 toss

Using the following formula

P(X=x) = nCr a^r * b ^ (n-r)

Where n represents total number of toss = 3

r represents number of occurrence

a represents getting a head = ½

b represents probability of getting a tail = ½

1. For h = 0

P(h=0) = 3C0 * ½^0 * ½³

P(h=0) = 1 * 1 * ⅛

P(h=0) = ⅛

2. For h = 1

P(h=1) = 3C1 * ½^1 * ½²

P(h=1) = 3 * ½ * ¼

P(h=1) = ⅜

3. P(h=2) = 3C2 * ½² * ½^1

P(h=2) = 3 * ¼ * ½

P(h=2) = ⅜

4.P(h=3) = 3C3 * ½³ * ½^0

P(h=0) = 1 * ⅛ * 1

P(h=0) = ⅛

It should be noted that when X is -5, h is either 1 or 2 and P(X) = ⅜

When X is -1, h is either 0 or 3 and P(X) = ⅛

The probability distribution function of X is as follows

Values of X || P(x)

-5 || ⅜

1 || ⅛

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Step-by-step explanation:

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