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Mars2501 [29]
3 years ago
7

Help me pls find this answer I doubt mine is right ​

Mathematics
1 answer:
Gnom [1K]3 years ago
8 0

Answer:

whats the question?

comment so i can help:)

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A Statistics class is estimating the mean height of all female students at their college. They collect a random sample of 36 fem
Galina-37 [17]

Answer:

d. ( 63.9, 66.7)

Therefore at 90% confidence interval (a,b) = ( 63.9, 66.7)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = 65.3

Standard deviation r = 5.2

Number of samples n = 36

Confidence interval = 90%

z(at 90% confidence) = 1.645

Substituting the values we have;

65.3+/-1.645(5.2/√36)

65.3+/-1.645(0.866667)

65.3+/-1.42567

65.3+/-1.4

= ( 63.9, 66.7)

Therefore at 90% confidence interval (a,b) = ( 63.9, 66.7)

4 0
4 years ago
Find the length of arc PQ in circle R to the nearest hundredth.
Sindrei [870]

Answer:

Step-by-step explanation:

7 0
3 years ago
Help with this it’s due today.!!!!
Leya [2.2K]

Answer:

Right answer is 10 I hope it's helps you

8 0
3 years ago
Question 5
Sedaia [141]
I think the answer is 48 hope this helps
7 0
4 years ago
Read 2 more answers
We know that the length of time required for a student to complete a particular aptitude test has a normal distribution with a m
Reptile [31]

Answer:

0.6832 = 68.32% probability that a given student will complete the test in more than 35 minutes but less than 43 minutes

Step-by-step explanation:

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.

Mean of 41.0 minutes and a variance of 3.4 minutes.

This means that \mu = 41, \sigma = 3.4

What is the probability, rounded to four decimal places, that a given student will complete the test in more than 35 minutes but less than 43 minutes?

This is the p-value of Z when X = 43 subtracted by the p-value of Z when X = 35.

X = 43

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

Z = \frac{43 - 41}{3.4}

Z = 0.59

Z = 0.59 has a p-value of 0.7224

X = 35

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

Z = \frac{35 - 41}{3.4}

Z = -1.76

Z = -1.76 has a p-value of 0.0392

0.7224 - 0.0392 = 0.6832

0.6832 = 68.32% probability that a given student will complete the test in more than 35 minutes but less than 43 minutes

4 0
3 years ago
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