Question

# Assume that thermometer reading are normally distributed with a mean of 0°C and a standard deviation...

Assume that thermometer reading are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A thermometer is randomly selected and tested. (a) Find the probability of a reading less than -2.75. (b) Find the probability of a reading greater than 2.33 (c) Find the probability of a reading between -2.87 and 1.34 (d) Find the 96th percentile

mean = 0

sd = 1

(a) P(x < -2.75)

P(z < -2.75) = 1- P(z < 2.75) = 1- 0.997 = 0.003

P(x < -2.75)  = 0.003

(b) P(x > 2.33)

P(z > 2.33) = 1- P(z < 2.33) = 1- 0.9901 = 0.0099

P(x > 2.33) = 0.0099

(c) P(-2.87 < x < 1.34)

P(-2.87 < z < 1.34) =  P( z < 1.34) -  P( z < - 2.87)

P( z < 1.34) = 0.9099

P( z < - 2.87) = 1- P( z < 2.87) = 1- 0.9979 = 0.0021

P(-2.87 < z < 1.34) = 0.9078

(d) 96th percentile

P(Z < z ) = 0.96

z = 1.751

x = 1.751

96th percentile; x = 1.751

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