Mathematics, 28.10.2020 16:50 snlawson9053

A survey found that women's heights are normally distributed with mean 64.2 in. and standard deviation 4.2 in. The survey also found that men's heights are normally distributed with mean 70.1 in. and standard deviation 4.1 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 57 in. and a maximum of 63 in. Complete parts (a) and (b) below

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Answer from: kylea97

A survey found that women's heights are normally distributed with mean 64.2 in. and standard deviation 4.2 in. The survey also found that men's heights are normally distributed with mean 70.1 in. and standard deviation 4.1 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 57 in. and a maximum of 63 in. Complete parts (a) and (b) below

a) find the percentage of women meeting the height requirement

b) find the percentage of men meeting the height requirement

P(57 < X < 63) = 34.4%

P(57 < X < 63) = 4.096%

Step-by-step explanation:

From the question we are told that

The mean of women's height is $\mu_ 1 = 64.2$

The standard deviation is $\sigma_1 = 4.2 \ in$

The mean of men' height is $\mu_2 = 70.1 \ in$

The standard deviation of men's height is $\sigma_2 = 4.1 \ in$

The minimum height requirement is a = 57 in

The maximum height requirement is b = 63 in

Generally the percentage of women meeting the height requirement is mathematically represented as

$P(57 < X < 63) = P(\frac{ 57 - 64.2}{ 4.2} < \frac{X - \mu_1}{\sigma_1} < \frac{ 63 - 64.2}{ 4.2} )$

$\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )$

P(57 < X < 63) = P( -1.7143 < Z

=> P(57 < X < 63) = P ( Z

Generally from the z table the probability of ( Z <-0.2857) and (Z < -1.7143) is

P( Z <-0.2857) = 0.38755

and

P(Z < -1.7143) = 0.043237

P(57 < X < 63) = 0.38755 - 0.043237

=> P(57 < X < 63) = 0.3443

Converting to percentage

P(57 < X < 63) = 0.3443 * 100 = 34.4%

Generally the percentage of men meeting the height requirement is mathematically represented as

$P(57 < X < 63) = P(\frac{ 57 - 70.1}{ 4.1} < \frac{X - \mu_1}{\sigma_1} < \frac{ 63 - 70.1}{ 4.1} )$

$\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )$

P(57 < X < 63) = P( -3.1951 < Z <-1.7317)

=> P(57 < X < 63) = P ( Z

Generally from the z table the probability of ( Z <-0.2857) and (Z < -1.7143) is

P( Z <-1.7317) = 0.041663

and

P(Z < -3.1951 ) = 0.00069891

P(57 < X < 63) = 0.041663 - 0.00069891

=> P(57 < X < 63) = 0.04096

Converting to percentage

P(57 < X < 63) = 0.04096 * 100 = 4.096%

Answer from: Quest

answer: yes, he is correct

step-by-step explanation: 10x-1=59

1-1=0 59+1=60

60/10=x=6

Answer from: Quest

What are the options

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