6 158-67 71 Totals 47o 2. The time it takes for babies to learn to walk isknown

Question

6 158-67 71 Totals 47o 2. The time it takes for babies to learn to walk isknown to be normally distributed with mean 9.5 months and standard deviation of 1 months. What percent of babies learn to walk: a) between 8 and 11 months. (inclusive) B more than 11 months. 5 points 5 points 5 points 5 points c) less than 8 months d) If the longest 1.15% are considered to be "late bloomers", how many months constitutes a "late bloomer" e) If the shortest 1.15% are considered as "ahead of their time", how many months constitutes being "ahead of their time" 5 point

Explanation / Answer

Given: µ = 9.5, = 1

To find the probability, we need to find the Z scores first.

Z = (X - µ)/ [/n]. Since n = 1, and = 1, Z = (X - µ)

(a) For P (8 < X < 11) = P(X < 11) – P(X < 8)

For P( X < 11)

Z = (11 – 9.5) = 1.5

The probability for P(X < 11) from the normal distribution tables is = 0.9332

For P( X < 8)

Z = (8 – 9.5) = -1.5

The probability for P(X < 8) from the normal distribution tables is = 0.0668

Therefore the required probability is 0.9332 – 0.0668 = 0.8664

(b) For P (X > 11) = 1 - P (X < 11), as the normal tables give us the left tailed probability only.

For P( X < 11)

Z = (11 – 9.5) = 1.5

The probability for P(X < 11) from the normal distribution tables is = 0.9332

Therefore the required probability = 1 – 0.9332 = 0.0668

(c) For P( X < 8)

Z = (8 – 9.5) = -1.5

The required probability from the normal distribution tables is = 0.0668

(d) P[Z = (X - µ)] > 1.5% = 0.015

Therefore P[Z = (X - µ)] < 1.5% = 1 - 0.015 = 0.985

The Z score at 0.985 is 2.17

Therefore X - 9.5 = 2.17 or X = 11.67 months

(e) P[Z = (X - µ)] < 1.5% = 0.015

The Z score at 0.015 is -2.17

Therefore X - 9.5 = -2.17 or X = 7.33 months

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