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Sveta_85 [38]
3 years ago
12

The times a fire department takes to arrive at the scene of an emergency are normally distributed with a mean of 6 minutes and a

standard deviation of 1 minute. For about what percent of emergencies does the fire department arrive at the scene in between 4 minutes and 8 minutes ?
Mathematics
1 answer:
Kitty [74]3 years ago
8 0

Answer:

Step-by-step explanation:

Let x be the random variable representing the times a fire department takes to arrive at the scene of an emergency. Since the population mean and population standard deviation are known, we would apply the formula,

z = (x - µ)/σ

Where

x = sample mean

µ = population mean

σ = standard deviation

From the information given,

µ = 6 minutes

σ = 1 minute

the probability that fire department arrives at the scene in case of an emergency between 4 minutes and 8 minutes is expressed as

P(4 ≤ x ≤ 8)

For x = 4,

z = (4 - 6)/1 = - 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.023

For x = 8

z = (8 - 6)/1 = 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.98

Therefore,

P(4 ≤ x ≤ 8) = 0.98 - 0.23 = 0.75

The percent of emergencies that the fire department arrive at the scene in between 4 minutes and 8 minutes is

0.75 × 100 = 75%

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2.35%

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According to the 68-95-99.7 rule, 95% of the data is comprised within two standard deviations of the mean (39-20 to 39+20 months), while 99.7% of the data is comprised within two standard deviations of the mean (39-30 to 39+30 months).

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8 0
3 years ago
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White raven [17]
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3 years ago
The Federal Government is stepping up efforts to reduce average response times of fire departments to fire calls. Nationwide, th
TiliK225 [7]

Answer: in order to qualify, the time must be at least 15.93 minutes

Step-by-step explanation:

Let x be a random variable representing the average response times of fire departments to fire calls. Since it follows a normal distribution, we will determine the z score by applying the formula,

z = (x - µ)/σ

Where

x = sample mean

µ = population mean

σ = population standard deviation

From the information given,

µ = 12.8 minutes

σ = 3.7 minutes

in order to qualify for the special safety award, the probability value would be to the right of the z score corresponding to 80%

The z score corresponding to 80% on the normal distribution table is 0.845

Therefore,

0.845 = (x - 12.8)/3.7

0.845 × 3.7 = x - 12.8

3.13 = x - 12.8

x = 12.8 + 3.13 = 15.93

7 0
3 years ago
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