A random sample is drawn from a population with mean μ = 66 and standard deviation σ = 5.5. use table 1.

1 answer:

7 0

A random sample is drawn from a population with mean μ = 66 and standard deviation σ = 5.5. use table 1.
a. is the sampling distribution of the sample mean with n = 16 and n = 36 normally distributed? yes, both the sample means will have a normal distribution. no, both the sample means will not have a normal distribution. no, only the sample mean with n = 16 will have a normal distribution. no, only the sample mean with n = 36 will have a normal distribution.
b. can you use the standard normal distribution to calculate the probability that the sample mean falls between 66 and 68 for both sample sizes? yes, for both the sample sizes, standard normal distribution could be used. no, for both the sample sizes, standard normal distribution could not be used. no, only for the sample size with n = 16, standard normal distribution could be used. no, only for the sample size with n = 36, standard normal distribution could be used.
c. calculate the probability that the sample mean falls between 66 and 68 for n = 36. (round intermediate calculations to 4 decimal places, "z" value to 2 decimal places, and final answer to 4 decimal places.)

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A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

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Answer:

(a) Probability that 2 or fewer will withdraw is 0.2061.

(b) Probability that exactly 4 will withdraw is 0.2182.

(c) Probability that more than 3 will withdraw is 0.5886.

(d) The expected number of withdrawals is 4.

Step-by-step explanation:

We are given that a university found that 20% of its students withdraw without completing the introductory statistics course.

Assume that 20 students registered for the course.

The above situation can be represented through binomial distribution;

$P(X =r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,......$

where, n = number of trials (samples) taken = 20 students

r = number of success

p = probability of success which in our question is probability

that students withdraw without completing the introductory

statistics course, i.e; p = 20%

Let X = Number of students withdraw without completing the introductory statistics course

So, X ~ Binom(n = 20 , p = 0.20)

(a) Probability that 2 or fewer will withdraw is given by = P(X $\leq$ 2)

P(X $\leq$ 2) = P(X = 0) + P(X = 1) + P(X = 2)

= $\binom{20}{0} \times 0.20^{0} \times (1-0.20)^{20-0}+ \binom{20}{1} \times 0.20^{1} \times (1-0.20)^{20-1}+ \binom{20}{2} \times 0.20^{2} \times (1-0.20)^{20-2}$

= $1 \times1 \times 0.80^{20}+ 20 \times 0.20^{1} \times 0.80^{19}+ 190\times 0.20^{2} \times 0.80^{18}$

= 0.2061

(b) Probability that exactly 4 will withdraw is given by = P(X = 4)

P(X = 4) = $\binom{20}{4} \times 0.20^{4} \times (1-0.20)^{20-4}$

= $4845\times 0.20^{4} \times 0.80^{16}$

= 0.2182

(c) Probability that more than 3 will withdraw is given by = P(X > 3)

P(X > 3) = 1 - P(X $\leq$ 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3)

= $1-(\binom{20}{0} \times 0.20^{0} \times (1-0.20)^{20-0}+ \binom{20}{1} \times 0.20^{1} \times (1-0.20)^{20-1}+ \binom{20}{2} \times 0.20^{2} \times (1-0.20)^{20-2}+\binom{20}{3} \times 0.20^{3} \times (1-0.20)^{20-3})$