If you save $125 each week for one year, how much will you be able to save? There are 52 weeks in one year.

MakcuM [25]

2,541 times 27 is: 68, 607. The equation is written like this: 2,541 x 27 = 68,607.

Hope I helped!

- Debbie

8 0

3 years ago

Thefirstassignmentinastatisticalcomputingclass involves running a short program. If past experience indicates that 40% of all st

Akimi4 [234]

Answer:

(a) P(X $\geq$ 25) = 0.0985

(b) P(15 $\leq$ X $\leq$ 25) = 0.8836

Step-by-step explanation:

We are given that past experience indicates that 40% of all students will make no programming errors. Also, a sample of class of 50 students is given.

Firstly, the above situation can be represented through binomial distribution, i.e.;

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

where, n = number of samples taken = 50

r = number of success

p = probability of success, i.e. 40%

Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.

So, Let X = No. of making no programming errors

Mean of X, $\mu$ = $n \times p$ = $50 \times 0.40$ = 20

Standard deviation of X, $\sigma$ = $\sqrt{np(1-p)}$ = $\sqrt{50 \times 0.40 \times (1-0.40)}$ = 3.5

So, X ~ N( $\mu = 20, \sigma^{2} = 3.5^{2}$ )

Now, the z score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

(a) Probability that in a class of 50 students at least 25 will make no errors is given by = P(X $\geq$ 25) = P(X > 24.5) ---- using continuity correction

P(X > 24.5) = P( $\frac{X-\mu}{\sigma}$ > $\frac{24.5-20}{3.5}$ ) = P(Z > 1.29) = 1 - P(Z $\leq$ 1.29)

= 1 - 0.90147 = 0.0985

(b) Probability that in a class of 50 students between 15 and 25 (inclusive) will make no errors = P(15 $\leq$ X $\leq$ 25) = P(X $\leq$ 25) - P(X < 15)

P(X $\leq$ 25) = P(X < 25.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{25.5-20}{3.5}$ ) = P(Z < 1.57) = 0.94179

P(X < 15) = P(X < 14.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{14.5-20}{3.5}$ ) = P(Z < -1.57) = 1 - P(Z $\leq$ 1.57)

= 1 - 0.94179 = 0.05821

Therefore, P(15 $\leq$ X $\leq$ 25) = 0.94179 - 0.05821 = 0.8836 .

4 0

3 years ago

A graduate student majoring in linguistics is interested in studying the number of students in her college who are bilingual. Of

Eddi Din [679]

Answer:

48.41% probability that 17 or fewer of them are bilingual.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .