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JulijaS [17]
3 years ago
10

he temperature at any random location in a kiln used in the manufacture of bricks is normally distributed with a mean of 925 and

a standard deviation of 60 degrees. If bricks are fired at a temperature above 1150, they will crack and must be disposed of. If the bricks are placed randomly throughout the kiln, the proportion of bricks that crack during the firing process is closest to
Mathematics
1 answer:
Leviafan [203]3 years ago
5 0

Answer:

The proportion of bricks that crack during the firing process is 0.0001 = 0.01%.

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 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.

Mean of 925 and a standard deviation of 60 degrees.

This means that \mu = 925, \sigma = 60

The proportion of bricks that crack during the firing process is closest to

This is 1 subtracted by the pvalue of Z when X = 1150. So

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

Z = \frac{1150 - 925}{60}

Z = 3.75

Z = 3.75 has a pvalue of 0.9999

1 - 0.9999 = 0.0001

The proportion of bricks that crack during the firing process is 0.0001 = 0.01%.

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One stats class consists of 52 women and 28 men. Assume the average exam score on Exam 1 was 74 (σ = 10.43; assume the whole cla
Svetllana [295]

Answer:

(A) What is the z- score of the sample mean?

The z- score of the sample mean is 0.0959

(B) Is this sample significantly different from the population?

No; at 0.05 alpha level (95% confidence) and (n-1 =79) degrees of freedom, the sample mean is NOT significantly different from the population mean.

Step -by- step explanation:

(A) To find the z- score of the sample mean,

X = 75 which is the raw score

¶ = 74 which is the population mean

S. D. = 10.43 which is the population standard deviation of/from the mean

Z = [X-¶] ÷ S. D.

Z = [75-74] ÷ 10.43 = 0.0959

Hence, the sample raw score of 75 is only 0.0959 standard deviations from the population mean. [This is close to the population mean value].

(B) To test for whether this sample is significantly different from the population, use the One Sample T- test. This parametric test compares the sample mean to the given population mean.

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S. E. = 16/√80 = 16/8.94 = 1.789

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Setting up the hypotheses,

Null hypothesis: Sample is not significantly different from population

Alternative hypothesis: Sample is significantly different from population

Having gotten T- cal, T- tab is found thus:

The Critical (Table) t value is found using

- a specific alpha or confidence level

- (n - 1) degrees of freedom; where n is the total number of observations or items in the population

- the standard t- distribution table

Alpha level = 0.05

1 - (0.05 ÷ 2) = 0.975

Checking the column of 0.975 on the t table and tracing it down to the row with 79 degrees of freedom;

The critical t value is 1.990

Since T- cal < T- tab (0.559 < 1.990), refute the alternative hypothesis and accept the null hypothesis.

Hence, with 95% confidence, it is derived that the sample is not significantly different from the population.

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