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A brand of water-softener salt comes in bags marked “net weight 18kg”. The
company that packages the salt claims that the bags contain an average of 18kg of
salt and that the standard deviation of the weight of the bag is 0.68kg. Assume that
the weight of the bags is
.05.
P = 0.012
z = = -2.33
μ=18
0.68
n = 10
Question
Consider the scenario where we randomly selected 10 bags of water
softener salt. If we were now interested in bags weighing both above and below
that claimed by the company.
a) Would you consider this evidence against the company’s claim?
b) In general, what
consider evidence against the company’s claim?
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- = 222 mg/dLi and Total cholesterol levels for middle aged men (aged 55 – 64) follow approximately a normal distribution with mean, µ standard deviation, o = 37 mg/dLi. Those with a total cholesterol of 240 mg/dLi or higher have "high" cholesterol (and have twice the risk of coronary heart disease as someone with a total cholesterol less than 200 mg/dLi) Find the Z-score (standard score) for an observation of 290 mg/dl.arrow_forwardAssume that the readings on the thermometers are normally distributed with a mean of 0° and standard deviation of 1.00°C. Assume 2.1% of the thermometers are rejected because they have readings that are too high and another 2.1% are rejected because they have readings that are too low. Draw a sketch and find the two readings that are cutoff values separating the rejected thermometers from the others.arrow_forward
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