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Nikitich [7]
4 years ago
11

Television viewing reached a new high when the Nielsen Company reported a mean daily viewing time of 8.35 hours per household (U

SA Today, November 11, 2009). Use a normal probability distribution with a standard deviation of 2.5 hours to answer the following questions about daily television viewing per household.
a. What is the probability that a household views television between 5 and 10 hours a day?
Mathematics
1 answer:
pogonyaev4 years ago
6 0

Answer:

Probability of viewing television between 5 to 10 hr is  0.6553

Step-by-step explanation:

Given data:

total time period of viewing of per household = 8.35 hr

standard deviation = 2.5 hr

Probability between 5 to 10 hr can be obtained as

= P(Between 5 and 10 hours)

= P(\frac{(5 - 8.35)}{2.5} < z < \frac{(10 - 8.35)}{2.5})

= P(-1.34 < z < 0.66)

= P(z < 0.66) - P(z < -1.34)

determine z value from standard z table

= 0.7454 - 0.0901

= 0.6553

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If a TV cost $3960.00 and toes on sale during the following week with a 25% discount, what will be the total cost of the TV? but
Free_Kalibri [48]

we have the price of TV is $3960.if we round off to the nearest tens we get same 3960 because we have 0 at the end which means lesser than 5 so 60 will be the nearest ten.

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7 0
3 years ago
If 10% of men are bald, what is the probability that fewer than 100 in a random sample of 818 men are bald? (Answers must be in
Greeley [361]

Answer: the probability that fewer than 100 in a random sample of 818 men are bald is 0.9830

Step-by-step explanation:

Given that;

p = 10% = 0.1

so let q = 1 - p = 1 - 0.1 = 0.9

n = 818

μ = np = 818 × 0.1 = 81.8

α = √(npq) = √( 818 × 0.1 × 0.9 ) = √73.62 = 8.58

Now to find P( x < 100)

we say;

Z = (X-μ / α) = ((100-81.8) / 8.58) = 18.2 / 8.58 = 2.12

P(x<100) = P(z < 2.12)

from z-score table

P(z < 2.12) = 0.9830

Therefore the probability that fewer than 100 in a random sample of 818 men are bald is 0.9830

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3 years ago
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mafiozo [28]

Answer:

254,000,00

I hope this helps

6 0
3 years ago
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