Mathematics, 13.08.2020 23:01, davidb1113

The population of men at UMBC has a mean height of 69 inches with a standard deviation of 4 inches. The women at UMBC have a mean height of 65 inches with a standard deviation of 3 inches. A sample of 50 men and 40 women is selected. What is the probability that the sample mean of men heights is more than 5 inches greater than the sample mean of women heights

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The probability that the sample mean of men heights is more than 5 inches greater than the sample mean of women heights is 0.0885.

Step-by-step explanation:

We are given that the population of men at UMBC has a mean height of 69 inches with a standard deviation of 4 inches. The women at UMBC have a mean height of 65 inches with a standard deviation of 3 inches.

A sample of 50 men and 40 women is selected.

The z-score probability distribution for the two-sample normal distribution is given by;

Z = $\frac{(\bar X_M-\bar X_W)-(\mu_M-\mu_W)}{\sqrt{\frac{\sigma_M^{2} }{n_M}+\frac{\sigma_W^{2} }{n_W} } }$ ~ N(0,1)

where, $\mu_M$ = population mean height of men at UMBC = 69 inches

$\mu_W$ = population mean height of women at UMBC = 65 inches

$\sigma_M$ = standard deviation of men at UMBC = 4 inches

$\sigma_M$ = standard deviation of women at UMBC = 3 inches

n_M = sample of men = 50

n_W = sample of women = 40

Now, the probability that the sample mean of men heights is more than 5 inches greater than the sample mean of women heights is given by = P( $\bar X_M-\bar X_W$ > 5 inches)

P( $\bar X_M-\bar X_W$ > 5 inches) = P( $\frac{(\bar X_M-\bar X_W)-(\mu_M-\mu_W)}{\sqrt{\frac{\sigma_M^{2} }{n_M}+\frac{\sigma_W^{2} }{n_W} } }$ > $\frac{(5)-(69-65)}{\sqrt{\frac{4^{2} }{50}+\frac{3^{2} }{40} } }$ ) = P(Z > 1.35)

= 1 - P(Z $\leq$ 1.35) = 1 - 0.9115 = 0.0885

The above probability is calculated by looking at the value of x = 1.35 in the z table which has an area of 0.9115.

answered: Guest

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.

answered: Guest

there would only be 50 odd numbers between 1 and 100 so therefor there would be 50 even so that would only be 50 odd numbers.

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