Question

In: Statistics and Probability

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 5.3...

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 5.3 years, and standard deviation of 1.7 years. If 16 items are picked at random, 2% of the time their mean life will be less than how many years? Give your answer to one decimal place.

A political candidate has asked you to conduct a poll to determine what percentage of people support her.
If the candidate only wants a 10% margin of error at a 98% confidence level, what size of sample is needed?
The political candidate will need to sample___ people.

A political candidate has asked you to conduct a poll to determine what percentage of people support her.
If the candidate only wants a 10% margin of error at a 98% confidence level, what size of sample is needed?
The political candidate will need to sample___ people.

You measure 38 textbooks' weights, and find they have a mean weight of 32 ounces. Assume the population standard deviation is 7.4 ounces. Based on this, construct a 90% confidence interval for the true population mean textbook weight.
Give your answers as decimals, to two places
I am 90% confident that the mean weight of textbooks is between___ and___ ounces.

A researcher studying the lifespan of a certain species of bacteria. A preliminary sample of 30 bacteria reveals a sample mean of ¯x=76x¯=76 hours with a standard deviation of s=6.8s=6.8 hours. He would like to estimate the mean lifespan for this species of bacteria to within a margin of error of 0.5 hours at a 99% level of confidence.
What sample size should you gather to achieve a 0.5 hour margin of error?
He would need to sample___ bacteria.

Solutions

Expert Solution

Solution:-

1) Given that mean = 5.3, sd = 1.7, n = 16,


P(X <= x0) = 0.02, for Z = -2.054

X = mean + Z*s/sqrt(n)
= 5.3 - (2.054*1.7/sqrt(16))
= 4.427
= 4.4 (rounded)

------------------------

2) for Assume that p = 0.5, E = 0.10, 98% Confidence interval for Z = 2.33

n = (Z/E)^2*p*q = (2.33/0.10)^2*0.5*0.5 = 135.722 = 136

The political candidate will need to sample 136 people.

----------------------

4. Given that n = 38, mean = 32, sd = 7.4, 90% Confidence interval

I am 90% confident that the mean weight of textbooks is between 30.03 and ,33.98 ounces.

Note : 2 and 3 sums are same



  


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