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enyata [817]
3 years ago
8

A blueberry plant produces about 26 blueberries per day. If the farmer picked 472 blueberries, how many days did the plants prod

uce blueberries?
Mathematics
1 answer:
Arte-miy333 [17]3 years ago
8 0

Answer:

About 18 days.  

Step-by-step explanation:

The problem states that a blueberry plant produces about 26 blueberries per day.  So, our answer will be somewhat of an estimate for the amount of days the plants produced blueberries based on the number picked.  In order to solve for the number of days, we can set up equivalent ratios (fractions):

\frac{blueberries}{day}=\frac{26}{1}=\frac{472}{d} where 'd' = number of days.

We can cross-multiply and divide in order to find the amount of days:

\frac{26d}{26}=\frac{472}{26}

So, d ≈ 18 days.

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Answer:

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f=2560

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A sample size 25 is picked up at random from a population which is normally
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Answer:

a) P(X < 99) = 0.2033.

b) P(98 < X < 100) = 0.4525

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 100 and variance of 36.

This means that \mu = 100, \sigma = \sqrt{36} = 6

Sample of 25:

This means that n = 25, s = \frac{6}{\sqrt{25}} = 1.2

(a) P(X<99)

This is the pvalue of Z when X = 99. So

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

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{99 - 100}{1.2}

Z = -0.83

Z = -0.83 has a pvalue of 0.2033. So

P(X < 99) = 0.2033.

b) P(98 < X < 100)

This is the pvalue of Z when X = 100 subtracted by the pvalue of Z when X = 98. So

X = 100

Z = \frac{X - \mu}{s}

Z = \frac{100 - 100}{1.2}

Z = 0

Z = 0 has a pvalue of 0.5

X = 98

Z = \frac{X - \mu}{s}

Z = \frac{98 - 100}{1.2}

Z = -1.67

Z = -1.67 has a pvalue of 0.0475

0.5 - 0.0475 = 0.4525

So

P(98 < X < 100) = 0.4525

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I guess I was help full for u

plz follow me for more updates

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