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valina [46]
3 years ago
5

PLEASE HELP ME !! PLEASE !!

Mathematics
1 answer:
Yakvenalex [24]3 years ago
6 0
I think that it is a ok
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Consider a normal distribution with mean 23 and standard deviation 7. What is the probability a value selected at random from th
Trava [24]

Answer:

0.5 = 50% probability a value selected at random from this distribution is greater than 23

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

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

In this problem, we have that:

\mu = 23, \sigma = 7

What is the probability a value selected at random from this distribution is greater than 23?

This is 1 subtracted by the pvalue of Z when X = 23. So

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

Z = \frac{23 - 23}{7}

Z = 0

Z = 0 has a pvalue of 0.5

0.5 = 50% probability a value selected at random from this distribution is greater than 23

5 0
3 years ago
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Please help me my grade is on the line
NemiM [27]
Answer shown and explanation above! Hope this helps!

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3 years ago
The length of time a person takes to decide which shoes to purchase is normally distributed with a mean of 8.21 minutes and a st
MaRussiya [10]

Correct answer is: P(x<6) is 0.123 and it is usual.

Solution:-

Given that the time a person takes to decide which shoes to purchase follows normal distribution. Which has mean = 8.21 minutes and standard deviation 1.90

Then probability of individual takes less than 6 minutes is

P(X<6) = P(z

           = P(z

           = 0.1230

Typically we say an event with a probability less than 5% is unusual.

But here P(X<6) = 0.123 is greater than 5% hence this is usual.

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Find a common multiplier then divide ex. 21/27 = 7/9 
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