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motikmotik
2 years ago
15

Use the appropriate normal distribution to approximate the resulting binomial distributions. A convenience store owner claims th

at 55% of the people buying from her store, on a certain day of the week, buy coffee during their visit. A random sample of 35 customers is made. If the store owner's claim is correct, what is the probability that fewer than 24 customers in the sample buy coffee during their visit on that certain day of the week?
Mathematics
1 answer:
Naddika [18.5K]2 years ago
6 0

Answer:

0.9256

Step-by-step explanation:

Given that a convenience store owner claims that 55% of the people buying from her store, on a certain day of the week, buy coffee during their visit

Let X be the number of customers who buy from her store, on a certain day of the week, buy coffee during their visit

X is Binomial (35, 0.55)

since each customer is independent of the other and there are two outcomes.

By approximation to normal we find that both np and nq are >5.

So X can be approximated to normal with mean = np = 19.25

and std dev = \sqrt{npq} \\=2.943

Required probability = prob that fewer than 24 customers in the sample buy coffee during their visit on that certain day of the week

= P(X (after effecting continuity correction)

= 0.9256

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blondinia [14]

Answer:

0.75 = 1-\frac{1}{k^2}

If we solve for k we can do this:

\frac{1}{k^2}= 1-0.75=0.25

\frac{1}{0.25}= k^2

k^2 =4

k =\pm 2

So then we have at last 75% of the data withitn two deviations from the mean so the limits are:

Lower = \mu -2\sigma = 514- 2*40=434

Upper = \mu +2\sigma = 514 + 2*40=594

Step-by-step explanation:

We don't know the distribution for the scores. But we know the following properties:

\mu = 514 , \sigma =40

For this case we can use the Chebysev theorem who states that "At least 1 -\frac{1}{k^2} of the values lies between \mu -k\sigma and \mu +k\sigma"

And we need the boundaries on which we expect at least 75% of the scores. If we use the Chebysev rule we have this:

0.75 = 1-\frac{1}{k^2}

If we solve for k we can do this:

\frac{1}{k^2}= 1-0.75=0.25

\frac{1}{0.25}= k^2

k^2 =4

k =\pm 2

So then we have at last 75% of the data withitn two deviations from the mean so the limits are:

Lower = \mu -2\sigma = 514- 2*40=434

Upper = \mu +2\sigma = 514 + 2*40=594

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Find the interest due to the bank on a loan of $1000 at 7.5% for 280 days
Lera25 [3.4K]

Answer:

The interest is \$57.53

Step-by-step explanation:

we know that

The simple interest formula is equal to

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