Question

10.02 5. About 38% of students participate in a community volunteer program outside of school. If 40 students are selected at random, find each of the following. 2MBRO (14 points) a) The mean and standard deviation of this binomial distribution. meb vididongo wil oldi b) The probability that exactly 18 students participate in this program using binomial formula. c) The probability that at least 15 students participate in this program using normal approximation Sole n tilidadonna 6. Birth weights are normal distributed with mean of 3152.1 grams and a standard deviation of 6934 grams. (14 points) a) Ir birth weight is randomly selected, find the probability that it is less than 2986.7 grams. b) If 25 birth weights are randomly selected, find the probability that their mean weight is more than 3607.5 grams. old man vill din bosbesichanol c) If 25 birth weights are randomly selected, find the probability that their mean weight is between 2558.4 and 3607.5 grams. sul islam totul al dolot blod i borimo oli el s ono

Answer 1

5).sample size (n) =40

p = probability of students participating in the program = 0.38

so, X ~ bin (40 , 0.38).

a). so, we have,

$mean = np=(40*0.38)=15.2$

$sd=\sqrt{npq}=\sqrt{40*0.38*0.62}=3.07$

b).the probability that exactly 18 students participating in the program be:-

$=P(X=18)$

$=_{18}^{40}\textrm{C}*0.38^{18}*0.62^{(40-18)}$

= 0.0838

c). here, np = 15.2 >5

nq = 40*0.62 =24.8 >5

so, we can use normal approximation.

so, now X ~ N ( 15.2 , 3.07)

the probability that at least 15 students will participate in the program be:-

$=P(X\geq15)$

$=P(X>14.5)$

$=P(\frac{X-15.2}{3.07}>\frac{14.5-15.2}{3.07})$

$=P(z>-0.23)$

$=1-\Phi(-0.23)$

$=\Phi(0.23)$

= 0.5910[ from standard normal table ] 6).X: birth weight

X ~ N (3152.1 , 693.4)

so, $\mu$ = 3152.1 , $\sigma$ = 693.4

a). the probability that the birth weight is less than 2986.7 grams be:-

$P(X<2986.7)$

$=P(\frac{X-3152.1}{693.4}<\frac{2986.7-3152.1}{693.4})$

$=P(z<-0.24)$

$=\Phi(-0.24)$

$=1-\Phi(0.24)$

$=1-0.5948$[ from standard normal table ]

= 0.4052

SOME CALCULATIONS FOR b and c :-

we have taken a sample of size 25 from the normal population.

sample size (n) = 25

so, the mean weight will follow normal distribution with,

mean = $\mu$ = 3152.1

$sd(s)=\frac{\sigma}{\sqrt{n}}=\frac{693.4}{\sqrt{25}}=138.68$

b). the probability that their mean weight is more than 3607.5 grams be:-

$P(X>3607.5)$

$=P(\frac{X-3152.1}{138.68}>\frac{3607.5-3152.1}{138.68})$

$=P(z>3.28)$

$=1-\Phi(3.28)$

$=1-0.9995$ [ from standard normal table ]

= 0.0005

c).the probability that their weight is between 2558.4 grams and 3607.5 grams be:-

$P(2258.4<X<3607.5)$

$P(-6.44<z<3.28)$

$=\Phi(3.28)-\Phi(-6.44)$

= 0.9995[ from standard normal table ]

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