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stellarik [79]
3 years ago
11

The Precision Scientific Instrument Company manufactures thermometers that are supposed to give readings of 0 degrees Celcius at

the freezing point of water. Tests on a large sample of these thermometers reveal that at the freezing point of water, some give readings below 0 degrees celcius (denoted by negative numbers) and some give readings above 0 degrees celcius (denoted by positive numbers).
Assume that the mean reading is 0 degrees celcius and the standard deviation of the readings is 1.00 degrees celcius. Also assume that the frequency distribution of errors closely resembles the normal distribution. A thermometer is randomly selected and tested. Find the temperature reading corresponding to the given information.

If 7% of the thermometers are rejected because they have readings that are too low, but all other thermometers are acceptable, find the temperature that separates the rejected thermometers from the others.
I know the answer, but do not how to solve the problem. answ-->-1.48

Find the z-value with a left tail of 7%:

invNorm(0.07) = -1.48

Find the termperature deviation that corresponds to that z-value:
x = z*s + u
x = -1.48*1 + 0 = -1.48
Mathematics
1 answer:
enot [183]3 years ago
6 0

Answer:

Step-by-step explanation:

Given that reading on thermometer is N (0,1)

7% of the thermometers are rejected because they have readings that are too low, but all other thermometers are acceptable

This implies that we have to find out the 7th percentile.

i.e. Z score = -1.475

Hence the temperature for cutoff is below -1.475 degrees.

the temperature reading corresponding to the given information.=-1.475 deg

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Answer:

There appears to be a difference between the pulse rates from samples of adult males and females. The average pulse rate of males is 63.75 while that of females is 75.875.

Coefficient of Variation (male) = 33.44%

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Step-by-step explanation:

Mean (male) = \frac{sum of x}{n} =\frac{1020}{16} =63.75

Mean (female) = \frac{sum of y}{n} = \frac{1214}{16} = 75.875

Standard Deviation (male) = \sqrt{\frac{sum(x-63.75)^2}{16-1} } =21.315

Standard Deviation (female) = \sqrt{\frac{sum(y-75.875)^2}{16-1} } = 21.512

Coefficient of Variation  = \frac{Std dev}{mean} *100

For male, CV = \frac{21.315}{63.75} *100=33.44%%

For female, CV = \frac{21.512}{75.875} *100=28.35%

Comparison : Pulse rates of males has more variability than that of females since it has a greater coefficient of variation.

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