The quotient of 5 plus d and 12 minus w

assume that when adults with smartphones are randomly selected 15 use them in meetings or classes if 15 adult smartphones are ra

Tanzania [10]

Answer:

The probability that at least 4 of them use their smartphones is 0.1773.

Step-by-step explanation:

We are given that when adults with smartphones are randomly selected 15% use them in meetings or classes.

Also, 15 adult smartphones are randomly selected.

Let X = <em>Number of adults who use their smartphones</em>

The above situation can be represented through the binomial distribution;

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

where, n = number of trials (samples) taken = 15 adult smartphones

r = number of success = at least 4

p = probability of success which in our question is the % of adults

who use them in meetings or classes, i.e. 15%.

So, X ~ Binom(n = 15, p = 0.15)

Now, the probability that at least 4 of them use their smartphones is given by = P(X $\geq$ 4)

P(X $\geq$ 4) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3)

= $1- \binom{15}{0}\times 0.15^{0} \times (1-0.15)^{15-0}-\binom{15}{1}\times 0.15^{1} \times (1-0.15)^{15-1}-\binom{15}{2}\times 0.15^{2} \times (1-0.15)^{15-2}-\binom{15}{3}\times 0.15^{3} \times (1-0.15)^{15-3}$

= $1- (1\times 1\times 0.85^{15})-(15\times 0.15^{1} \times 0.85^{14})-(105 \times 0.15^{2} \times 0.85^{13})-(455 \times 0.15^{3} \times 0.85^{12})$

= <u>0.1773</u>

3 0

3 years ago

An experiment to investigate the survival time in hours of an electronic component consists of placing the parts in a test cell

mixer [17]

Answer:

a) The sample mean is of 49 and the sample standard deviation is of 11.7.

b) The range of the true mean at 90% confidence level is of 9.62 hours.

c) The prediction interval, at a 90% confidence level, of it's failure time is between 39.38 hours and 58.62 hours.

Step-by-step explanation:

Question a:

Sample mean:

$\overline{x} = \frac{34+40+46+49+61+64}{6} = 49$

Sample standard deviation:

$s = sqrt{\frac{(34-49)^2+(40-49)^2+(46-49)^2+(49-49)^2+(61-49)^2+(64-49)^2}{5}} = 11.7$

The sample mean is of 49 and the sample standard deviation is of 11.7.

b)Determine the range of the true mean at 90% confidence level.

We have to find the margin of error of the confidence interval. Since we have the standard deviation for the sample, the t-distribution is used.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 6 - 1 = 5

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 5 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$ . So we have T = 2.0.150