See the attached file.
Assume that women’s height are normally distributed with a meangiven by =63.4 in, and a standard deviationgiven by =2.7in.
If 1 woman is randomly selected, find the probabilitythat her height is less than 64 in.
If 46 women are randomly selected find the probability that they have a mean height less than 64 in.
Cans of a certain beverage are labeled to indicate that they contain 8oz. The amounts in a sample of cans are measured and the sample statisticsare n=44 and x=8.04oz. If the beverage cans are filled so that µ=8.00 oz (as labeled) and the population standard deviations is σ=0.107 oz (based on the sample results), find the probability that a sample of 44 cans will have a mean of 8.04 oz or greater. Do these results suggest that the beverage cans are filled with an amount greater than 8.00 oz?
The probability that a sample of 44 cans will have a mean of 8.04 oz or greater, given that µ =8.00 and σ =0.107, is____
(a) With n=11 and p=0.6, find the binomialprobability P(5) by using a binomial table. (b) If np ≥ 5 and nq ≥ 5, also estimate the indicated probability by using the normal distributionas an approximation to the binomial; if np < 5 or nq < 5, then state the normal approximation cannot be used.
Find the probability by using a binomial probability table.
P(5)=___(round 3 decimal places as needed)
Assume the readings on thermometers are normally distributed with a mean of 0 C and a standard deviation of 1.00 C. Find the probability that a randomly selected thermometer reads less than 0.13 and draw a sketch of the region.
a) If the random variablez is the standard normal score and a>0, is it true that P (z<-a)= P(z>a)? Why or why not?
b) Find the z-score for the standard normal distribution where Area=0.32 in the left trail.
Please see the attached file.
1. Assume that women’s height are normally distributed with a mean given by =63.4 in, and a standard deviation given by =2.7in.
a. If 1 woman is randomly selected, find the probability that her height is less than 64 in.
Let X be the height of one woman, then P(X<64)=P(Z<(64-63.4)/2.7)=P(Z<0.22)=0.5871 from z table.
b. If 46 women are randomly selected find the probability that they have a mean height less than 64 in.
Let Xbar be the mean height of 46 women, then P(Xbar<64)=
P(Z<(64-63.4)*sqrt(46)/2.7)=P(Z<1.51)=0.9345 from z table.
2. Cans of a certain beverage are labeled to indicate …
The solution gived detailed steps on solving a series of questions on some distributions: normal, t and binomial. All formula and calculations are shown and explained.